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Solution to the practice set of ladder race LV1 (all)
2022-07-06 11:25:00 【%xiao Q】
L1-001 Hello World (5 branch )
A simple question
#include <stdio.h>
int main()
{
printf("Hello World!\n");
return 0;
}
L1-002 Print hourglass (20 branch )
This question mainly includes 3 A pit
- No extra symbols ,0 Also output
- The space after the symbol does not need to be filled , Otherwise, the format error will be reported , But the space before the symbol still needs to be printed
Reference code ：
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n;
char ch;
cin >> n >> ch;
int x = sqrt((n + 1) / 2); // Calculate how many layers
// Output inverted triangle
for(int i = 1; i <= x; i++)
{
for(int j = 1; j < i; j++) cout << ' ';
for(int j = 0; j < 2 * (x - i + 1) - 1; j++) cout << ch;
cout << endl;
}
// Output integer triangle
for(int i = 2; i <= x; i++)
{
for(int j = 0; j < x - i; j++) cout << ' ';
for(int j = 0; j < 2 * i - 1; j++) cout << ch;
cout << endl;
}
int ans = n - 2 * x * x + 1;
cout << ans << endl; // No extra symbols ,0 Also output
return 0;
}
L1-003 Single digit statistics (15 branch )
The data range of this question is not large , So just use an array to record the number of occurrences of all numbers .
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1010;
int st[N];
int main()
{
string s;
cin >> s;
for(int i = 0; i < s.size(); i++) st[s[i] - '0']++;
for(int i = 0; i <= 1000; i++)
if(st[i]) printf("%d:%d\n", i, st[i]);
return 0;
}
L1-004 Calculate Celsius (5 branch )
According to the meaning of the question
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
cin >> n;
printf("Celsius = %d\n", 5 * (n - 32) / 9);
return 0;
}
L1-005 Test seat number (15 branch )
This question can be used map Hash , First deposit all map in , Then just look for it directly .
Reference code ：
#include <iostream>
#include <unordered_map>
using namespace std;
unordered_map<int, pair<string, int>> mp;
int main()
{
string s;
int n, m, T;
cin >> T;
for(int i = 0; i < T; i++)
{
cin >> s >> n >> m;
mp[n] = {
s, m};
}
cin >> T;
while(T--)
{
cin >> n;
cout << mp[n].first << ' ' << mp[n].second << endl;
}
return 0;
}
L1-006 Continuous factor (20 branch )
This question requires the number of longest continuous factors , And output the smallest sequence of continuous factors .
So what can violence enumerate , So how to enumerate ？
First of all, we can know from the data range , The maximum length will not be greater than 10, So let's enumerate the length （ From big to small ）, Then enumerate the first number of continuous factors at the beginning , Then we start by judging the length and the number , Can it be established , If set up , That is the result we need .
Reference code ：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
int n;
cin >> n;
int maxn = sqrt(n);
for(int len = 10; len >= 1; len--)
for(int start = 2; start <= maxn; start++) // Less than maxn Because ,sqrt(n) * (sqrt(n) + 1) Greater than n
{
int t = 1;
for(int i = start; i <= start + len - 1; i++)
t *= i;
if(n % t == 0)
{
printf("%d\n%d", len, start);
for(int i = start + 1; i <= start + len - 1; i++)
printf("*%d", i);
return 0;
}
}
printf("1\n%d", n);
return 0;
}
L1-007 Read numbers (10 branch )
A simple question , Directly in the code
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
string st[] = {
"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
int main()
{
string s;
cin >> s;
for(int i = 0; i < s.size(); i++)
{
if(s[i] == '-') cout << "fu";
else cout << st[s[i] - '0'];
if(i != s.size() - 1) cout << ' '; // Control the last space
}
cout << endl;
return 0;
}
L1-008 Sum of integral segments (10 branch )
Directly according to the meaning of the simulation can be
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b, sum = 0, t = 0;
cin >> a >> b;
for(int i = a; i <= b; i++)
{
sum += i;
t++;
printf("%5d", i);
if(t == 5)
{
t = 0;
printf("\n");
}
}
if(t != 0) printf("\n");
printf("Sum = %d\n", sum);
return 0;
}
L1-009 N Sum the numbers (20 branch )
This problem is very stupid , First of all, you have to deal with your output , Then you have to add them up .
Processing input , here , We directly store two arrays ,a The array stores molecules ,b The array has a denominator ,
Then add each one in turn , Add up , Directly divide and add , And then make an appointment .
Reference code ：
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int ans1 = 0, ans2 = 0;
void slove(int a, int b)
{
// Here is the first one
if(!ans1 && !ans2)
{
ans1 = a, ans2 = b;
return ;
}
// Divide and add
int x2 = ans2 * b;
int x1 = ans1 * b + a * ans2;
// Apposition
int t = __gcd(x1, x2); // Find the greatest common divisor
ans1 = x1 / t, ans2 = x2 / t;
}
int main()
{
int n;
string s;
cin >> n;
while(n--)
{
int a = 0, b = 0;
cin >> s;
int t = 0;
bool flag = false; // Used to judge the negative sign later
for(int i = 0; i < s.size(); i++)
{
if(s[i] == '-') flag = true;
else if(s[i] == '/')
{
if(flag) a = t * -1;
else a = t;
flag = false;
t = 0;
}
else
t = t * 10 + s[i] - '0';
}
if(flag) b = t * -1;
else b = t;
slove(a, b);
}
if(ans2 == 1) cout << ans1 << endl;
else
{
if(ans1 / ans2) cout << ans1 / ans2 << ' ';
cout << ans1 - ans2 * (ans1 / ans2) << '/' << ans2 << endl;
}
return 0;
}
L1-010 Compare the size (10 branch )
This question is relatively simple , I just put the code in .
Reference code ：
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int a[3];
int main()
{
cin >> a[0] >> a[1] >> a[2];
sort(a, a + 3); // Sort
printf("%d->%d->%d\n", a[0], a[1], a[2]);
return 0;
}
L1-011 A-B (20 branch )
The first B String use map Mark the , Then traverse A String it , During traversal, judge whether to output .
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
unordered_map<char, int> mp;
int main()
{
string s1, s2;
getline(cin, s1); // Read in a whole line , You can read in spaces , At the end of a carriage return
getline(cin, s2);
for(int i = 0; i < s2.size(); i++)
mp[s2[i]] = 1;
for(int i = 0; i < s1.size(); i++)
if(!mp.count(s1[i])) cout << s1[i];
puts("");
return 0;
}
L1-012 Calculate the index (5 branch )
A water problem
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
cin >> n;
printf("2^%d = %d\n", n, 1 << n); // 1 << n = 2 ^ n
return 0;
}
L1-013 Calculate factorial sum (10 branch )
direct 2 Layer loops solve problems .
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int ans = 0;
for(int i = 1; i<= n; i++)
{
int sum = 1;
for(int j = 1; j <= i; j++)
sum *= j;
ans += sum;
}
cout << ans << endl;
return 0;
}
L1-014 Simple questions (5 branch )
Big water
#include <iostream>
using namespace std;
int main()
{
cout << "This is a simple problem." << endl;
return 0;
}
L1-015 Draw squares with Obama (15 branch )
Go straight to the code .
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
char ch;
cin >> n >> ch;
// To round
int m = (n + 1) / 2;
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
cout << ch;
cout << endl;
}
return 0;
}
L1-016 Check your ID card (15 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int w[] = {
7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2}; // Save the corresponding weight
char st[] = {
'1', '0', 'X', '9', '8', '7', '6', '5', '4', '3', '2'}; // Save the corresponding check code
int main()
{
int n;
cin >> n;
string s;
int t = 0;
for(int i = 0; i < n; i++)
{
cin >> s;
int sum = 0;
bool flag = true;
for(int i = 0; i < s.size() - 1; i++)
{
if(s[i] >= '0' && s[i] <= '9')
sum += w[i] * (s[i] - '0');
else
{
flag = false;
break;
}
}
char p = st[sum % 11];
if(!flag || p != s[s.size() - 1])
{
cout << s << endl;
t++;
}
}
if(t == 0) puts("All passed");
return 0;
}
L1-017 How many (15 branch )
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
string s;
cin >> s;
int cnt = 0;
for(int i = 0; i < s.size(); i++)
if(s[i] == '2') cnt++;
int n = s.size();
if(s[0] == '-') n--;
double ans = 1.0 * cnt / n;
if(s[0] == '-') ans *= 1.5;
int t = s[s.size() - 1] - '0';
if(t % 2 == 0) ans *= 2;
printf("%.2f%\n", ans * 100);
return 0;
}
L1-018 Big Ben (10 branch )
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b;
scanf("%d:%d", &a, &b);
if(a >= 0 && a <= 11 || a == 12 && b == 0)
printf("Only %02d:%02d. Too early to Dang.", a, b); // %02d： Less than two zeros
else
{
a -= 12;
string s = "Dang";
if(b > 0) a++;
for(int i = 0; i < a; i++) cout << s;
puts("");
}
return 0;
}
L1-019 Who first down (15 branch )
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n, A, B;
int a, b, c, d;
cin >> A >> B >> n;
int cnt_a = 0, cnt_b = 0;
for(int i = 0; i < n; i++)
{
cin >> a >> b >> c >> d;
if(b == d) continue; // Two people draw the same , It is impossible to win or lose , Go straight to the next cycle
int t = a + c;
if(t == b) cnt_a++; // a A penalty
else if(t == d) cnt_b++; // b A penalty
// a Fall down
if(cnt_a > A)
{
cout << "A" << endl;
cout << cnt_b << endl;;
break;
}
// b Fall down
if(cnt_b > B)
{
cout << "B" << endl;
cout << cnt_a << endl;
break;
}
}
return 0;
}
L1-020 Too handsome to have any friends (20 branch )
This question can be used directly map Make a mark , Then in the process of query , Also used map Make a mark , Prevent duplicate output 2 The same person .
Reference code ：
#include <iostream>
#include <cstdio>
#include <unordered_map>
using namespace std;
unordered_map<string, int> mp;
unordered_map<string, int> st;
int main()
{
int T;
cin >> T;
while(T--)
{
int n;
cin >> n;
string s;
for(int i = 0; i < n; i++)
{
cin >> s;
if(n > 1) mp[s]++;
}
}
cin >> T;
int t = 0;
while(T--)
{
string s;
cin >> s;
if(st.count(s)) continue;
st[s]++;
if(!mp.count(s))
{
if(t != 0) cout << ' ';
t++;
cout << s;
}
}
if(t == 0) printf("No one is handsome");
puts("");
return 0;
}
L1-021 Say the important thing three times (5 branch )
Go straight to the code ：
#include <iostream>
using namespace std;
int main()
{
for(int i = 0; i < 3; i++)
cout << "I'm gonna WIN!" << endl;
return 0;
}
L1-022 Parity separation (10 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
cin >> n;
int t1 = 0, t2 = 0;
for(int i = 0; i < n; i++)
{
int x;
cin >> x;
if(x % 2) t1++;
else t2++;
}
cout << t1 << ' ' << t2 << endl;
return 0;
}
L1-023 Output GPLT (20 branch )
It can be used 1 Save an array 4 Number of letters , Then in the “GPLT” Output in the order of
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int a[4];
int main()
{
string s;
cin >> s;
for(int i = 0; i < s.size(); i++)
{
if(s[i] == 'G' || s[i] == 'g') a[0]++;
else if(s[i] == 'P' || s[i] == 'p') a[1]++;
else if(s[i] == 'L' || s[i] == 'l') a[2]++;
else if(s[i] == 'T' || s[i] == 't') a[3]++;
}
while(a[0] || a[1] || a[2] || a[3])
{
if(a[0])
{
cout << 'G';
a[0]--;
}
if(a[1])
{
cout << 'P';
a[1]--;
}
if(a[2])
{
cout << 'L';
a[2]--;
}
if(a[3])
{
cout << 'T';
a[3]--;
}
}
cout << endl;
return 0;
}
L1-024 The day after tomorrow (5 branch )
Go straight to the code ：
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int ans = (n + 2) % 7;
if(ans == 0) cout << 7 << endl; // by 0 Time is Sunday
else cout << ans << endl;
return 0;
}
L1-025 Positive integer A+B (15 branch )
Generally speaking, this question , There are several points to pay attention to ：
- The first space is A,B Separation of
- A May be an empty string
So we can read it directly in a whole line , And then we're doing segmentation , Divide into A,B, Then make a judgment
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
string s1, s2, s;
getline(cin, s);
bool flag = true;
for(int i = 0; i < s.size(); i++)
{
if(flag && s[i] == ' ')
{
flag = false;
continue;
}
if(flag) s1 += s[i];
else s2 += s[i];
}
// cout << s1 << endl << s2 << endl;
bool flag1 = true, flag2 = true;
for(int i = 0; i < s1.size(); i++)
if(s1[i] < '0' || s1[i] > '9') flag1 = false;
for(int i = 0; i < s2.size(); i++)
if(s2[i] < '0' || s2[i] > '9') flag2 = false;
if(s1.size() == 0) flag1 = false;
int t1 = 0, t2 = 0;
if(flag1)
{
if(s1.size() > 4)
{
cout << "?";
flag1 = false;
}
else
{
for(int i = 0; i < s1.size(); i++)
t1 = t1 * 10 + s1[i] - '0';
if(t1 > 1000 || t1 == 0)
{
cout << "?";
flag1 = false;
}
else cout << t1;
}
}
else cout << "?";
cout << " + ";
if(flag2)
{
if(s2.size() > 4)
{
cout << "?";
flag2 = false;
}
else
{
for(int i = 0; i < s2.size(); i++)
t2 = t2 * 10 + s2[i] - '0';
if(t2 > 1000 || t2 == 0)
{
cout << "?";
flag2 = false;
}
else cout << t2;
}
}
else cout << "?";
cout << " = ";
if(flag1 && flag2) cout << t1 + t2;
else cout << "?";
cout << endl;
return 0;
}
L1-026 I Love GPLT (5 branch )
Go straight to the code ：
#include <iostream>
using namespace std;
int main()
{
string s = "I Love GPLT";
for(int i = 0; i < s.size(); i++)
cout << s[i] << endl;
return 0;
}
L1-027 lease (20 branch )
We can save all the different numbers , Then sort , Then open an array to save and convert its subscript and value .
Then traverse the phone number .
Reference code ：
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 200;
int ar[N];
int arr[N], n;
int index[N], m;
bool st[N];
bool cmp(int x, int y)
{
return x > y;
}
int main()
{
string s;
cin >> s;
for(int i = 0; i < s.size(); i++)
{
int t = s[i] - '0';
if(!st[t])
arr[n++] = t;
st[t] = true;
}
sort(arr, arr + n, cmp); // Sort
printf("int[] arr = new int[]{%d", arr[0]);
for(int i = 1; i < n; i++)
printf(",%d", arr[i]);
printf("};\n");
// Exchange subscripts and values
for(int i = 0; i < n; i++)
{
int t = arr[i];
ar[t] = i;
}
for(int i = 0; i < s.size(); i++)
{
int t = s[i] - '0';
index[m++] = ar[t];
}
printf("int[] index = new int[]{%d", index[0]);
for(int i = 1; i < m; i++)
printf(",%d", index[i]);
printf("};\n");
return 0;
}
L1-028 Judgement primes (10 branch )
Judge directly by trial and division
Reference code ：
#include <iostream>
#include <cstdio>
using namespace std;
bool judge(int x)
{
if(x == 1) return false;
for(int i = 2; i <= x / i; i++)
if(x % i == 0) return false;
return true;
}
int main()
{
int T;
cin >> T;
while(T--)
{
int x;
cin >> x;
if(judge(x))
puts("Yes");
else puts("No");
}
return 0;
}
L1-029 Is it too fat (5 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int h;
cin >> h;
double ans = (h - 100) * 0.9 * 2;
printf("%.1f\n", ans);
return 0;
}
L1-030 One gang, one (15 branch )
This question can be used 2 An array is saved for boys and girls , And then go through all the students , And find the matching students ,
Because all the students are entered in the order of ranking , So when saving, boys and girls also come in order , So we find matching students , You can find it directly at the end , Mark after finding , Prevent duplicate pairing .
Reference code ：
#include <iostream>
#include <cstdio>
#include <vector>
#include <unordered_map>
using namespace std;
typedef pair<int, string> PIS;
const int N = 55;
PIS s[N];
vector<string> boy; // Save boys
vector<string> girl; // Save girls
unordered_map<string, int> st;
int main()
{
int n;
cin >> n;
for(int i = 0; i < n; i++)
{
int a; string b;
cin >> a >> b;
s[i] = {
a, b};
if(a == 0) girl.push_back(b);
else boy.push_back(b);
}
int r1 = boy.size() - 1, r2 = girl.size() - 1;
for(int i = 0; i < n; i++)
{
if(st.count(s[i].second)) continue;
if(s[i].first == 0)
{
cout << s[i].second << ' ';
cout << boy[r2] << endl;
st[boy[r2]] = 1;
r2--;
}
else
{
cout << s[i].second << ' ';
cout << girl[r1] << endl;
st[girl[r1]] = 1;
r1--;
}
st[s[i].second] = 1;
}
return 0;
}
L1-031 Is it too fat (10 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
int T;
cin >> T;
while(T--)
{
int h, w;
cin >> h >> w;
double t = (h - 100) * 0.9 * 2;
if(abs(w - t) < t * 0.1) puts("You are wan mei!");
else if(w > t) puts("You are tai pang le!");
else puts("You are tai shou le!");
}
return 0;
}
L1-032 Left-pad (20 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
char ch;
string s;
cin >> n >> ch;
getchar();
getline(cin, s);
if(n <= s.size())
for(int i = s.size() - n; i < s.size(); i++)
cout << s[i];
else
{
for(int i = 0; i < n - s.size(); i++) cout << ch;
for(int i = 0; i < s.size(); i++) cout << s[i];
}
puts("");
return 0;
}
L1-033 Year of birth (15 branch )
This question needs attention , The different numbers happen to be n position
Reference code ：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int y, n;
bool st[10];
// How many digits are different in judgment
bool judge(int x)
{
memset(st, 0, sizeof st);
int cnt = 0;
for(int i = 0; i < 4; i++)
{
int t = x % 10;
if(!st[t]) cnt++;
st[t] = true;
x /= 10;
}
if(cnt == n) return true;
return false;
}
int main()
{
cin >> y >> n;
int t = 0;
for(int i = y; ; i++)
{
if(judge(i))
{
printf("%d %04d\n", t, i); // I don't remember enough 4 Place zero
break;
}
else t++;
}
return 0;
}
L1-034 give the thumbs-up (20 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1010;
int a[N];
int main()
{
int n;
cin >> n;
for(int i = 0; i < n; i++)
{
int m;
cin >> m;
for(int j = 0; j < m; j++)
{
int x;
cin >> x;
a[x]++;
}
}
int ans, maxn = -1;
for(int i = 1; i <= 1000; i++)
{
if(a[i] >= maxn)
{
maxn = a[i];
ans = i;
}
}
cout << ans << ' ' << maxn << endl;
return 0;
}
L1-035 Valentine's Day (15 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
string a, b, s;
int main()
{
int t = 0;
while(1)
{
cin >> s;
if(s.size() == 1 && s[0] == '.') break;
t++;
if(t == 2) a = s;
else if(t == 14) b = s;
}
if(t < 2) puts("Momo... No one is for you ...");
else if(t < 14) cout << a << " is the only one for you..." << endl;
else cout << a << " and " << b << " are inviting you to dinner...\n";
return 0;
}
L1-036 A multiply B (5 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
cout << a * b << endl;
return 0;
}
L1-037 A Divide B (10 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
if(b == 0)
{
printf("%d/%d=Error\n", a, b);
return 0;
}
double ans = 1.0 * a / b;
if(b > 0)
printf("%d/%d=%.2f\n", a, b, ans);
else
printf("%d/(%d)=%.2f\n", a, b, ans);
return 0;
}
L1-038 The new world (5 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
printf("Hello World\nHello New World\n");
return 0;
}
L1-039 Antique typesetting (20 branch )
Go straight to the code ：
#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n;
string s;
cin >> n;
getchar();
getline(cin, s);
int m = s.size();
int t = ceil(1.0 * m / n); // Rounding up , Find out how many columns
// Output according to the meaning of the topic
for(int i = 0; i < n; i++)
{
for(int j = t - 1; j >= 0; j--)
{
int d = j * n + i;
if(d >= s.size()) cout << ' ';
else cout << s[d];
}
puts("");
}
return 0;
}
L1-040 Height difference of the best couple (10 branch )
Go straight to the code ：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int T;
cin >> T;
while(T--)
{
double x;
char ch;
cin >> ch >> x;
if(ch == 'F')
printf("%.2f\n", x * 1.09);
else
printf("%.2f\n", x / 1.09);
}
return 0;
}
L1-041 seek 250 (10 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int x, t = 0;
while(cin >> x)
{
t++;
if(x == 250)
{
cout << t << endl;
break;
}
}
return 0;
}
L1-042 Date formatting (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b, c;
scanf("%d-%d-%d", &a, &b, &c);
printf("%04d-%02d-%02d\n", c, a, b);
return 0;
}
L1-043 Reading Room (20 branch )
/* This topic has 3 One thing to note ： 1. If you borrow a book , Start from the last time you borrow books 2. After returning the book, you may still borrow it 3. Then the average here is rounded */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1010;
int st[N];
bool vis[N];
int main()
{
int n;
cin >> n;
int t = 0, cnt = 0, ans = 0;
while(1)
{
int id, a, b;
char ch;
scanf("%d %c %d:%d", &id, &ch, &a, &b);
if(id == 0)
{
t++;
double sum = 0;
if(cnt != 0)
sum = round(1.0 * ans / cnt); // To round
cout << cnt << ' ' << sum << endl;
memset(st, 0, sizeof st);
memset(vis, 0, sizeof vis);
cnt = ans = 0;
if(t == n) break;
continue;
}
if(ch == 'S')
{
st[id] = a * 60 + b;
vis[id] = true;
}
else
{
if(vis[id])
{
cnt++;
ans += a * 60 + b - st[id];
vis[id] = false;
}
}
}
return 0;
}
L1-044 Wynn (15 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int t;
cin >> t;
int cnt = 0;
while(1)
{
string s;
cin >> s;
if(s == "End") break;
if(cnt == t)
{
cout << s << endl;
cnt = 0;
}
else
{
cnt++;
if(s == "ChuiZi") puts("Bu");
else if(s == "JianDao") puts("ChuiZi");
else puts("JianDao");
}
}
return 0;
}
L1-045 The universe is invincible (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
string s;
cin >> s;
cout << "Hello " << s << endl;
return 0;
}
L1-046 Get rid of the bachelors (20 branch )
/* Ideas ： This problem can be solved by the principle of division , We can find another one t = 1, And then divide by x, The remainder in use is ,t % x Come and ride 10 + 1, Of course, we have to pay attention to the leading zero here . */
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int x;
cin >> x;
int t = 1, len = 1;
bool flag = false;
while(1)
{
if(t / x != 0) flag = true; // Prevent output leading zeros
if(flag) cout << t / x;
if(t % x == 0) break;
t %= x;
t = t * 10 + 1;
len++;
}
cout << ' ' << len << endl;
return 0;
}
L1-047 Pretend to sleep (10 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
cin >> n;
while(n--)
{
string s;
int a, b;
cin >> s >> a >> b;
// cout << s << ' ' << a << ' ' << b;
if(a < 15 || a > 20 || b < 50 || b > 70) cout << s << endl;
}
return 0;
}
L1-048 matrix A multiply B (15 branch )
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 110;
int a[N][N], b[N][N], ans[N][N];
int main()
{
int ra, ca, rb, cb;
cin >> ra >> ca;
for(int i = 0; i < ra; i++)
for(int j = 0; j < ca; j++) cin >> a[i][j];
cin >> rb >> cb;
for(int i = 0; i < rb; i++)
for(int j = 0; j < cb; j++) cin >> b[i][j];
if(ca != rb) printf("Error: %d != %d\n", ca, rb);
else
{
cout << ra << ' ' << cb << endl;
for(int i = 0; i < ra; i++)
{
for(int j = 0; j < cb; j++)
{
int sum = 0;
for(int k = 0; k < ca; k++) sum += a[i][k] * b[k][j];
if(j != 0) cout << ' ';
cout << sum;
}
puts("");
}
}
return 0;
}
L1-049 Seat allocation for the ladder race (20 branch )
/* There is one thing to pay attention to in this topic ： That is, if there is only one team , The number is from 1 Start , If the test point 2 error , Consider the situation of a team */
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
const int N = 110;
int a[N];
bool st[N];
vector<int> ans[N];
int main()
{
int n;
cin >> n;
for(int i = 0; i < n; i++) cin >> a[i];
int t = 0, cnt = 0;
if(n == 1) t = -1; // Judge the situation of a team
while(1)
{
if(cnt == n) break;
for(int i = 0; i < n; i++)
{
if(st[i]) continue;
if(ans[i].size() + 1 > a[i] * 10)
{
cnt++;
st[i] = true;
continue;
}
if(cnt == n - 1)
{
ans[i].push_back(t + 2);
t += 2;
}
else ans[i].push_back(++t);
}
}
for(int i = 0; i < n; i++)
{
printf("#%d\n", i + 1);
int t = 0;
for(int j = 0; j < ans[i].size(); j++)
{
t++;
if(t == 10)
{
t = 0;
cout << ans[i][j] << endl;
}
else cout << ans[i][j] << ' ';
}
}
return 0;
}
L1-050 Last but not least N A string (15 branch )
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int m = 0;
char s[15];
int main()
{
int L, n;
cin >> L >> n;
int t = pow(26, L) - n; // Find the number of positive digits
while(t)
{
int x = t % 26;
s[m++] = 'a' + x;
t /= 26;
}
// When the string is insufficient L position , Remember to fill in the lead 'a'
if(m < L)
{
for(int i = 0; i < L - m; i++) cout << 'a';
}
for(int i = m - 1; i >= 0; i--) cout << s[i];
cout << endl;
return 0;
}
L1-051 Discount (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
double ans = n * m / 10.0;
printf("%.2f\n", ans);
return 0;
}
L1-052 2018 We want to win (5 branch )
PHP（ The best language in the world , dog's head ）
2018
wo3 men2 yao4 ying2 !
L1-053 Electronic Wang (10 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
for(int i = 0; i < a + b; i++) cout << "Wang!";
cout << endl;
return 0;
}
L1-054 Good luck. (15 branch )
/* Precautions for this topic ： The reversal of this question is left and right , Turn it upside down */
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 110;
int n;
char ch;
string s[N];
int main()
{
cin >> ch >> n;
getchar();
for(int i = 0; i < n; i++) getline(cin, s[i]);
/* Judge whether the left and right sides are symmetrical , But I found that the data is more water , I got full marks without making up or down judgments （ But out of the correctness of the solution , It still gives up and down judgments ） */
bool flag = true;
for(int i = 0; i < n; i++)
{
int l = 0, r = s[i].size() - 1;
while(l < r)
{
if(s[i][l] != s[i][r])
{
flag = false;
break;
}
l++, r--;
}
if(!flag) break;
}
// Judge whether the top and bottom are symmetrical
bool flag0 = true;
if(flag) // If left-right asymmetry , Then there is no need to judge whether the top and bottom are symmetrical
{
for(int i = 0; i < n; i++)
{
int l = 0, r = n - 1;
while(l < r)
{
if(s[l][i] != s[r][i])
{
flag0 = false;
break;
}
l++, r--;
}
if(!flag0) break;
}
}
if(!flag || !flag0)
{
for(int i = 0; i < n; i++)
{
int l = 0, r = s[i].size() - 1;
while(l < r)
{
swap(s[i][l], s[i][r]);
l++, r--;
}
}
}
else puts("bu yong dao le");
for(int i = n - 1; i >= 0; i--)
{
for(int j = 0; j < s[i].size(); j++)
if(s[i][j] != ' ') cout << ch;
else cout << s[i][j];
puts("");
}
return 0;
}
L1-055 Who is the winner (10 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int pa, pb;
cin >> pa >> pb;
int a = 0, b = 0;
for(int i = 0; i < 3; i++)
{
int x;
cin >> x;
if(!x) a++;
else b++;
}
if(pa > pb)
{
if(b == 3) printf("The winner is b: %d + %d\n", pb, b);
else printf("The winner is a: %d + %d\n", pa, a);
}
else
{
if(a == 3) printf("The winner is a: %d + %d\n", pa, a);
else printf("The winner is b: %d + %d\n", pb, b);
}
return 0;
}
L1-056 Guess the number (20 branch )
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 1e4 + 10;
typedef pair<string, int> PSI; // You can baidu pair Usage of , Good use of
int n;
PSI a[N];
int main()
{
cin >> n;
int sum = 0;
for(int i = 0; i < n; i++)
{
string s;
int x;
cin >> s >> x;
sum += x;
a[i] = {
s, x};
}
int avg = sum / n / 2;
// Find the closest avg Number of numbers , And record the subscript
int t = -1, minn = 2e9;
for(int i = 0; i < n; i++)
{
int d = abs(a[i].second - avg);
if(d < minn)
{
t = i;
minn = d;
}
}
cout << avg << ' ' << a[t].first << endl;
return 0;
}
L1-057 PTA It makes me energetic (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
printf("PTA shi3 wo3 jing1 shen2 huan4 fa1 !\n");
return 0;
}
L1-058 6 flared up (15 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
string s;
getline(cin, s);
for(int i = 0; i < s.size(); i++)
{
if(s[i] != '6') cout << s[i];
else
{
int j = i;
while(s[j] == '6') j++;
if(j - i > 9) cout << 27;
else if(j - i > 3) cout << 9;
else
for(int k = 0; k < j - i; k++) cout << 6;
i = j - 1;
}
}
puts("");
return 0;
}
L1-059 Knock the stupid bell (20 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int T;
cin >> T;
getchar();
while(T--)
{
string s;
getline(cin, s);
// Judge whether it rhymes
bool flag1 = false, flag2 = false;
for(int i = 0; i < s.size(); i++)
{
if(s[i] == ',' && s[i - 3] == 'o' && s[i - 2] == 'n' && s[i - 1] == 'g') flag1 = true;
if(s[i] == '.' && s[i - 3] == 'o' && s[i - 2] == 'n' && s[i - 1] == 'g') flag2 = true;
}
if(flag1 && flag2)
{
// find 3 A space （ namely 3 word ）
int t = 0, idx;
for(int i = s.size() - 1; i >= 2; i--)
{
if(s[i] == ' ') t++;
if(t == 3)
{
idx = i;
break;
}
}
for(int i = 0; i <= idx; i++) cout << s[i];
cout << "qiao ben zhong.\n";
}
else puts("Skipped");
}
return 0;
}
L1-060 Psychological shadow area (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int x, y;
cin >> x >> y;
int ans = 5000 - 0.5 * x * y - 0.5 * (100 - x) * (100 - y) - (100 - x) * y;
cout << ans << endl;
return 0;
}
L1-061 The new fat formula (10 branch )
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
double x, y;
cin >> x >> y;
double ans = x / (pow(y, 2));
printf("%.1f\n", ans);
if(ans > 25) puts("PANG");
else puts("Hai Xing");
return 0;
}
L1-062 Lucky lottery (15 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int T;
cin >> T;
while(T--)
{
string s;
cin >> s;
int sum1 = 0, sum2 = 0;
for(int i = 0; i < 3; i++) sum1 += s[i] - '0';
for(int i = 3; i < 6; i++) sum2 += s[i] - '0';
if(sum1 == sum2) puts("You are lucky!");
else puts("Wish you good luck.");
}
return 0;
}
L1-063 Fish or meat (10 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int T;
cin >> T;
while(T--)
{
int sex, h, w;
cin >> sex >> h >> w;
if(sex)
{
if(h < 130) cout << "duo chi yu!";
else if(h > 130) cout << "ni li hai!";
else cout << "wan mei!";
if(w < 27) cout << " duo chi rou!";
else if(w > 27) cout << " shao chi rou!";
else cout << " wan mei!";
cout << endl;
}
else
{
if(h < 129) cout << "duo chi yu!";
else if(h > 129) cout << "ni li hai!";
else cout << "wan mei!";
if(w < 25) cout << " duo chi rou!";
else if(w > 25) cout << " shao chi rou!";
else cout << " wan mei!";
cout << endl;
}
}
return 0;
}
L1-064 With a valuation of 100 million AI Core code (20 branch )
Because my code did not get full marks , Stick the @ssf_cxdm Blogger's code ,（ This question is so difficult to adjust ）
#include<iostream>
#include<vector>
#include<string>
#include<sstream>
#include<cstdio>
using namespace std;
int main(){
int n;
scanf("%d",&n);
// Don't forget this sentence
getchar();
while(n--){
string s,str[1005],s1;
int cnt=0;
getline(cin,s);
cout<<s<<endl;
cout<<"AI:";
for(int i=0;i<s.size();i++){
if(isalnum(s[i])){
//isalnum(), Letters or Numbers
if(s[i]!='I')
s[i]=tolower(s[i]); //tolower()
}else{
// Divided into pieces
s.insert(i," "); //s.insert()
i++;
}
if(s[i]=='?')
s[i]='!';
}
stringstream ss(s); // Eliminate spaces
while(ss>>s1){
str[cnt++]=s1;
}
if(!isalnum(str[0][0]))
cout<<" ";
for(int i=0;i<cnt;i++){
if(!isalnum(str[i][0])){
// There is no space before punctuation
cout<<str[i];
}
else if(str[i]=="can"&&(i+1<cnt&&str[i+1]=="you")){
cout<<" I can";
i++;
}
else if(str[i]=="could"&&(i+1<cnt&&str[i+1]=="you")){
cout<<" I could";
i++;
}
else if(str[i]=="I"||str[i]=="me"){
cout<<" you";
}
else cout<<" "<<str[i]; // The first character is a letter or a number , You have to add a space in front
}
cout<<endl;
}
return 0;
}
L1-065 Nonsense code (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
cout << "Talk is cheap. Show me the code." << endl;
return 0;
}
L1-066 Cats are liquids (5 branch )
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a, b, c;
cin >> a >> b >> c;
cout << a * b * c << endl;
return 0;
}
L1-067 Roche limit (10 branch )
#include <iostream>
#include <cstdio>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
double a, b, c;
cin >> a >> b >> c;
if(b) a *= 1.26;
else a *= 2.455;
if(a < c)
printf("%.2f ^_^\n", a);
else
printf("%.2f T_T\n", a);
return 0;
}
L1-068 Harmonic mean (10 branch )
// A stupid question , Chinese understanding belongs to , Have a good understanding of ： Their harmonic mean is the reciprocal of their reciprocal arithmetic mean
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
int n;
cin >> n;
double sum = 0;
rep(i, 1, n)
{
double x;
cin >> x;
x = 1 / x;
sum += x;
}
printf("%.2f\n", 1 / (sum / n));
return 0;
}
L1-069 Tire pressure monitoring (15 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int a[4];
//int ans[4];
int main()
{
int x, y, maxn = -1;
reps(i, 0, 4)
{
cin >> a[i];
maxn = max(maxn, a[i]);
}
cin >> x >> y;
int t1 = 0, t2 = 0, ans1, ans2;
reps(i, 0, 4)
{
if(a[i] < x)
{
t1++;
ans1 = i + 1;
}
if(abs(a[i] - maxn) > y)
{
t2++;
ans2 = i + 1;
}
}
if(t1 == 0 && t2 == 0) puts("Normal");
else if(t1 == 1 && t2 == 0) printf("Warning: please check #%d!\n", ans1);
else if(t1 == 0 && t2 == 1) printf("Warning: please check #%d!\n", ans2);
else if(t1 == 1 && t2 == 1 && ans1 == ans2) printf("Warning: please check #%d!\n", ans1);
else puts("Warning: please check all the tires!");
return 0;
}
L1-070 Eat hot pot (15 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
string s;
int ans1 = 0, ans2, t = 0;
bool flag = true;
while(getline(cin, s))
{
if(s == ".") break;
t++;
// cout << t << endl;
if(s.find("chi1 huo3 guo1") != -1) ans1++;
if(ans1 == 1 && flag) ans2 = t, flag = false;
}
cout << t << endl;
if(ans1 == 0) puts("-_-#");
else cout << ans2 << ' ' << ans1 << endl;
return 0;
}
L1-071 Previous life archives (20 branch )
// This question can try to find the law , But you can also take a look at the law of full binary tree first
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
while(m--)
{
int t = n - 1;
string s;
cin >> s;
int ans = 1;
reps(i, 0, s.size())
{
if(s[i] == 'n') ans += 1 << t;
t--;
}
cout << ans << endl;
}
return 0;
}
L1-072 Scratch the lottery (20 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int st[] = {
0,0,0,0,0,0,10000,36,720,360,80,252,108,72,54,180,72,180,119,36,306,1080,144,1800,3600};
int a[5][5];
bool vis[10];
int main()
{
int x, y;
rep(i, 1, 3) // Macro definition , amount to 1~3 The cycle of , You can see the above define, The following questions may appear , You can learn a little bit about
rep(j, 1, 3)
{
cin >> a[i][j];
if(a[i][j] == 0) x = i, y = j;
vis[a[i][j]] = true;
}
rep(i, 1, 9)
if(!vis[i])
{
a[x][y] = i;
break;
}
rep(i, 1, 3)
{
cin >> x >> y;
cout << a[x][y] << endl;
}
int t, sum = 0;;
cin >> t;
if(t <= 3)
rep(j, 1, 3) sum += a[t][j];
else if(t <= 6)
rep(i, 1, 3) sum += a[i][t - 3];
else if(t == 7)
sum = a[1][1] + a[2][2] + a[3][3];
else sum = a[1][3] + a[2][2] + a[3][1];
cout << st[sum] << endl;
return 0;
}
L1-073 Man and God (5 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
printf("To iterate is human, to recurse divine.\n");
return 0;
}
L1-074 Two hours C Language (5 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
int a, b, c;
cin >> a >> b >> c;
cout << a - b * c << endl;
}
L1-075 Obsessive compulsive disorder (10 branch )
#include <cstdio>
#include <iostream>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
int main()
{
string s;
cin >> s;
if(s.size() == 4)
{
int t = (s[0] - '0') * 10 + s[1] - '0';
if(t < 22) printf("20%c%c-%c%c", s[0], s[1], s[2], s[3]);
else printf("19%c%c-%c%c", s[0], s[1], s[2], s[3]);
}
else
{
rep(i, 0, 3) printf("%c", s[i]);
printf("-%c%c", s[4], s[5]);
}
return 0;
}
L1-076 Price reduction reminder robot (10 branch )
#include <cstdio>
#include <iostream>
#include <vector>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
vector<double> ans;
int main()
{
int n, m;
cin >> n >> m;
while(n--)
{
double x;
cin >> x;
if(x < m) ans.push_back(x);
}
reps(i, 0, ans.size())
{
// cout << i << endl;
printf("On Sale! %.1f", ans[i]);
if(i != ans.size() - 1) puts("");
}
return 0;
}
L1-077 Big Ben's mood (15 branch )
#include <cstdio>
#include <iostream>
#include <vector>
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
using namespace std;
vector<int> b;
int a[24];
int main()
{
reps(i, 0, 24) cin >> a[i];
int t;
while(cin >> t)
{
if(t < 0 || t > 23) break;
b.push_back(t);
}
reps(i, 0, b.size())
{
printf("%d ", a[b[i]]);
if(a[b[i]] > 50) cout << "Yes";
else cout << "No";
if(i != b.size() - 1) cout << endl;
}
return 0;
}
L1-078 Teacher Ji's return (15 branch )
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include <unordered_map>
#define LL long long
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
#define pre(i, a, b) for(int i = b; i >= a; i--)
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
getchar();
// bool flag = false;
while(n--)
{
string s;
getline(cin, s);
if(s.find("qiandao") != -1 || s.find("easy") != -1) continue;
if(m == 0)
{
cout << s;
return 0;
}
m--;
}
if(m >= 0) cout << "Wo AK le";
return 0;
}
L1-079 The goodness of TIANTI race (20 branch )
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include <unordered_map>
#define LL long long
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
#define pre(i, a, b) for(int i = b; i >= a; i--)
using namespace std;
const int N = 1e6 + 10;
int st[N];
int main()
{
int n;
cin >> n;
int minn = 2e9, maxn = -1;
rep(i, 1, n)
{
int x;
cin >> x;
minn = min(minn, x);
maxn = max(maxn, x);
st[x]++;
}
cout << minn << ' ' << st[minn] << endl;
cout << maxn << ' ' << st[maxn];
return 0;
}
L1-080 Multiplication formula sequence (20 branch )
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include <unordered_map>
#define LL long long
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define reps(i, a, b) for(int i = a; i < b; i++)
#define pre(i, a, b) for(int i = b; i >= a; i--)
using namespace std;
vector<int> a;
int main()
{
int x, y, n;
cin >> x >> y >> n;
a.push_back(x); a.push_back(y);
int i = 0, j = 1;
while(a.size() <= n)
{
int t = a[i] * a[j];
if(t >= 10)
{
a.push_back(t / 10);
a.push_back(t % 10);
}
else a.push_back(t);
i++, j++;
}
rep(i, 0, n - 1)
{
cout << a[i];
if(i != n - 1) cout << ' ';
}
return 0;
}
Here we are ,LV1 It's all updated , If there is a problem , It can be put forward in the comments , Bloggers must answer when they see ！！！
It may be updated later LV2 Of ,,,haha.
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