Niuke Xiaobai monthly race 49

A

Sign in

After reading the string , Convert bits into numbers ,%3 Output

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include<unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
const int N = 1e4 + 5;
const int MOD = 1e9 + 7;

char s[3], t[3];
int main() {
    
	scanf("%s %s", s, t);
	int a = s[0] - '0' + t[0] - '0';
	a %= 3;
	int b = s[1] - '0' + t[1] - '0';
	b %= 3;
	int c = s[2] - '0' + t[2] - '0';
	c %= 3;
	printf("%d%d%d", a, b, c);
	return 0;
}

B

simulation

Now look , I feel it's troublesome to write
use s The array stores the gifts received and the corresponding colors （ Write too much , It leads to direct search , A waste of time , Should be assigned directly , Find it like this O(1)）
Then I didn't input a color , Just judge once
Because color judgment is cumulative , It means that the last thing we input is 3 2 1 , Judge the first 3 when , Just find a gift with three , Judge the second 2 When , To statistics meanwhile There are color 3 and 2 A gift from
If a gift is in a certain color judgment , Not meeting the conditions , Just take the first one of the gift colors The assignment is 1, When traversing, judge whether to skip

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include<unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
const int N = 205;
const int MOD = 1e9 + 7;

int n, m;
int s[N][N];

void solve() {
    
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++) {
    
		for (int j = 0; j < m; j++) {
    
			scanf("%d", &s[i][j]);
		}
	}

	int a, cnt = 0;
	for (int i = 0; i < m; i++) {
    
		scanf("%d", &a);
		for (int j = 0; j < n; j++) {
    
			bool flag = false;
			for (int k = 0; k < m; k++) {
    
				if (s[j][0] == -1) break;
				if (s[j][k] == a) {
    
					cnt++, flag = true;
					break;
				}
			}
			if (flag == false) s[j][0] = -1;
		}
		printf("%d ", cnt);
		cnt = 0;
	}
	
	printf("\n");
}

int main() {
    
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}

Bucket row

vis[], The first i Item , Contains color …, This is a direct assignment ,O(1) lookup
judge[] It's No i Is the gift still qualified
b[] Receive the color to find

First traverse the color , Inner traversal gift , Judge gifts one by one , If b The color of the array , This gift has , And also meet the conditions , Just ans++
otherwise judge = 1, It means that the conditions are not met , After traversing a color, output it

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include<unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
const int N = 205;
const ll MOD = 1e9 + 7;

int n, m, a;
int vis[N][N]; //  The first  i  Pieces of   goods   contain   Color  j
int judge[N]; //  The first  i  Pieces of   goods   Whether the conditions are still met 
int b[N];

void solve() {
    
	memset(vis, 0, sizeof(vis));
	memset(judge, 0, sizeof(judge));
	memset(b, 0, sizeof(b));
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++) {
    
			scanf("%d", &a);
			vis[i][a] = 1; //  The first  i  Item , There are color a
		}
		
	for (int i = 0; i < m; i++) scanf("%d", &b[i]);
	
	for (int j = 0; j < m; j++) {
    
		int ans = 0;
		for (int i = 0; i < n; i++) {
    
			if (vis[i][b[j]]) {
    
				if (!judge[i]) ans++;
			}
			else {
    
				judge[i] = 1;
			}
		}
		printf("%d%c", ans, (j == m - 1) ? '\n' : ' ');
	}

}

int main() {
    
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}

C

An operation + thinking

a_i & a_j ,& There are only two corresponding positions yes 1 When it comes to 1, therefore ,& The operation can only be original 1 The quantity remains the same , Or reduce , It's impossible to increase , therefore & The subsequent value must be less than or equal to ai and aj Of , And it is smaller than the smaller one of them
And then there is ,| operation , In fact Add 1 The operation of , Because one must be ai & ai Of , The result is ai, Everything else & value , It must be less than or equal to ai Of , all ai | Other values , The result is ai
therefore The final result is actually hold a1~an All the values XOR up

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include<unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
const int N = 5e5 + 5;
const int MOD = 1e9 + 7;

int n, a[N];
long long res = 0;
void solve() {
    
	scanf("%d", &n);
	res = 0;
	for (int i = 1; i <= n; i++) {
    
		scanf("%d", &a[i]);
		res ^= a[i];
	}
	printf("%lld\n", res);
}

int main() {
    
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}

D

Quadratic function image + Inverse element

From the quadratic function image, we can get , The maximum value is actually greater than 0 Part of ,S(b) - S(a)
Then cut off the quadratic function , Find the sum of each term separately , And then add it up

Inverse element part , Generally, there are pits in division , There are decimals in division , But it's divisible , I made an appointment , The title does not say that it is a downward division , So divide —> Multiplication , Using the inverse element
Topic mod + 1 Then it happens to be divisible 2 and 6 Of , So you can write conveniently

iv2 = (MOD + 1) / 2ll;
iv6 = (MOD + 1) / 6ll;

Remember to take the mold ,ans and Output S(b) - S(a)
Sure add MOD Then take the mold , Prevent overflow ;

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include<unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
const int N = 1e4 + 5;
const ll MOD = 998244353;
const ll iv2 = (MOD + 1) / 2ll;
const ll iv6 = (MOD + 1) / 6ll;
ll a, b;
ll S(int t) {
    
	ll ans1 = t * (t + 1ll) % MOD * (2ll * t + 1ll) % MOD * iv6 % MOD;
	ll ans2 = (a + b) * t % MOD * (t + 1ll) % MOD * iv2 % MOD;
	ll ans3 = a * b % MOD * t % MOD;
	ll ans = -ans1 + ans2 - ans3;
	ans = (ans % MOD + MOD) % MOD;
	return ans;
}

void solve() {
    
	scanf("%lld%lld", &a, &b);
	printf("%lld\n", (S(b) - S(a) + MOD) % MOD);
}

int main() {
    
	int T;
	scanf("%d", &T);
	while (T--)
		solve();
	return 0;
}