2021 robocom world robot developer competition - preliminary competition of higher vocational group

2021 RoboCom World robot developer competition - Preliminary competition of higher vocational group

Preface

During the competition, the teacher always said , You can't even compete in Higher Vocational Education .

The questions are simple , Just post the code directly , Some use C++ Written , Some use Python Written .

1. The robot says hello

AC Code

print("ni ye lai can jia RoboCom a?")

2. Face recognition

AC Code

#include<bits/stdc++.h>
using namespace std;
signed main(){
    
    int A,B,C,D,a,b,c;
    cin>>A>>B>>C>>D>>a>>b>>c;
    cout<<"Diff = "<<A-a<<", "<<B-b<<", "<<C-c<<"\n";
    if (abs(A-a) + abs(B-b) + abs(C-c) <= D)
        cout<<"Yes";
    else
        cout<<"No";
    return 0;
}

3. Monthly output

AC Code

#include<bits/stdc++.h>
using namespace std;
int n;
string s[13] = {
    "","January","February","March","April","May","June",
                "July","August","September","October","November","December"};
signed main(){
    
    while (cin>>n){
    
        if (n > 12 || n < 1)
            break;
        cout<<s[n]<<"\n";
    }
    cout<<"?";
    return 0;
}

4. A string of letters

AC Code

#include<bits/stdc++.h>
using namespace std;
signed main(){
    
    int n;cin>>n;
    while(n--){
    
        string s;cin>>s;bool f = true;
        for (int i = 0; i < s.length() - 1; ++i) {
    
            if (islower(s[i])){
    
                if (!(s[i + 1] == s[i] - 1 || toupper(s[i]) == s[i + 1])){
    
                    f = false;break;
                }
            }
            else
                if (!(tolower(s[i]) == s[i + 1] || s[i + 1] == s[i] + 1)){
    
                    f = false;break;
                }
        }
        if (f)
            cout<<"Y\n";
        else
            cout<<"N\n";
    }
    return 0;
}

5. Increase by one

AC Code

#include<bits/stdc++.h>
using namespace std;
#define int long long
bool check(int x){
    
    int k = 0,kk = x;
    while (kk){
    
        kk /= 10; k += 1;
    }
    if (k % 2 != 0)
        return false;
    k = pow(10,k / 2);
    if (x % k - 1 == x / k)
        return true;
    return false;
}
signed main(){
    
    int n;cin>>n;
    while(n--){
    
        int m,k = 0;
        cin>>m;
        if (check(m)){
    
            k += 1;
            if (ceil(sqrt(m)) * ceil(sqrt(m)) == m)
                k += 1;
        }
        cout<<k<<"\n";
    }
    return 0;
}

6. Answer sheet

A netizen posted an article on Sina Weibo saying ：“ My friend bought Ben Bohr X Heisenberg's physics boss is humanistic , Yes 300 A high number question . What is more perfect , To finish the question, paint the answer sheet according to the answer , Paint a QR code , Scan the QR code to see the classic , If you make a mistake, you won't see it ……” The legendary answer sheet is shown in the figure below ： If the answer is 4 An integer （ When the number of digits is insufficient, make up in the front 0）, Then the first two are abscissa , The last two are ordinates . If there are two questions in a question , Then the answer to the first question is abscissa , The answer to the second question is the ordinate . If the answer is a score , Then the molecule is the abscissa , The denominator is the ordinate .

Please help the reader fill in the answer card according to the answers .

AC Code

ch = [['.'] * 105 for i in range(105)]
n, m = map(int,input().split())
for i in range(m):
    s = input()
    if s.count(';'):
        a, b = map(int,s.split(';'))
        col = a
        row = n - b + 1
        ch[row][col] = '#'
    elif s.count('/'):
        a, b = map(int,s.split('/'))
        col = a
        row = n - b + 1
        ch[row][col] = '#'
    else:
        a = int(s)
        b = a % 100
        a //= 100
        col = a
        row = n - b + 1
        ch[row][col] = '#'
for i in range(1,n + 1):
    for j in range(1,n + 1):
        print(ch[i][j],end='')
    print()

7. Save the unlucky guy

AC Code

s = set()
n = int(input())
for i in range(n):
    s.add(input())
m = int(input())
for i in range(m):
    s.remove(input())
s = list(s)
s.sort(reverse=True)
for i in s:
    print(i)