J4A 宾果游戏

思路

依次判断横排，竖排，正对角线与反对角线是否均为 $1$，如果是，则输出 Yes，如果都不是，则输出 No。

复杂度

令 $n$ 为矩阵大小。

时间复杂度:

$O(n^2)$

空间复杂度:

$O(n^2)$

Code

#include <bits/stdc++.h>
using namespace std;
int a[10][10];
int main()
{
	int n = 5;
	for (int i = 1;i <= n;i++)
	{
		for (int j = 1;j <= n;j++)
		{
			cin >> a[i][j];
		}
	}
	int f = 0;
	for (int i = 1;i <= n;i++)
	{
		int s = 0;
		for (int j = 1;j <= n;j++) s = s + a[i][j];
		if (s == 5) f = 1;
	}
	for (int j = 1;j <= n;j++)
	{
		int s = 0;
		for (int i = 1;i <= n;i++) s = s + a[i][j];
		if (s == 5) f = 1;
	}
	int s = 0;
	for (int i = 1;i <= n;i++)
	{
		s = s + a[i][i];
	}
	if (s == 5) f = 1;
	s = 0;
	for (int i = 1;i <= n;i++)
	{
		s = s + a[i][n + 1 - i];
	}
	if (s == 5) f = 1;
	cout << (f == 1 ? "Yes" : "No");
}