判断上三角矩阵。输入第一行给出一个正整数T,为待测矩阵的个数。接下来给出T个矩阵的信息:每个矩阵信息的第一行给出一个不超过10的正整数n。随后n行,每行给出n个整数,其间以空格分隔。

输入样例:

2
3
1 2 3
0 4 5
0 0 6
2
1 0
-8 2

输出样例:

YES
NO


#include <iostream>
using namespace std;

int main(int argc, char *argv[])
{
	
	int t;
	cin >> t;
	int *n = new int[t];
	int ***a = new int **[t];
	for (int i = 0; i < t; i++)
	{
		cin >> n[i];
		a[i] = new int*[n[i]];
		for (int j = 0; j < n[i]; j++)
		{
			a[i][j] = new int[n[i]];	
			for (int k = 0; k < n[i]; k++)
			{
				cin >> a[i][j][k];
			}
		}
		
	}


	for (int i = 0; i < t; )
	{
		for (int j = 0; j < n[i]; j++)
		{
			for (int k = 0; k < j; k++)
			{
				if (a[i][j][k] != 0)
				{
					cout << "NO"<<endl;
					goto OUT;
				}
			}
		}
		cout << "YES"<<endl;
	OUT:
		i++;

	}

	for (int i = 0; i < t; i++)
	{
		
		for (int j = 0; j < t; j++)
		{
			delete[]a[i][j];
		}
		delete[]a[i];
	}
	delete[]a;


	


	return 0;
}


猜你喜欢

转载自blog.csdn.net/xbpalyer/article/details/88761962