Codeforces 821A-Okabe and Future Gadget Laboratory

2967 단어 Codeforces폭력
Okabe and Future Gadget Laboratory
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Okabe needs to renovate the Future Gadget Laboratory after he tried doing some crazy experiments! The lab is represented as an n byn square grid of integers. A good lab is defined as a lab in which every number not equal to 1 can be expressed as the sum of a number in the same row and a number in the same column. In other words, for every x, y such that 1 ≤ x, y ≤ n and ax, y ≠ 1, there should exist two indices s and t so that ax, y = ax, s + at, y, where ai, j denotes the integer in i-th row and j-th column.
Help Okabe determine whether a given lab is good!
Input
The first line of input contains the integer n (1 ≤ n ≤ 50) — the size of the lab.
The next n lines contain n space-separated integers denoting a row of the grid. The j-th integer in the i-th row is ai, j (1 ≤ ai, j ≤ 105).
Output
Print "Yes"if the given lab is good and "No"otherwise.
You can output each letter in upper or lower case.
Examples
input
3
1 1 2
2 3 1
6 4 1

output
Yes

input
3
1 5 2
1 1 1
1 2 3

output
No

Note
In the first sample test, the 6 in the bottom left corner is valid because it is the sum of the 2 above it and the 4 on the right. The same holds for every number not equal to 1 in this table, so the answer is "Yes".
In the second sample test, the 5 cannot be formed as the sum of an integer in the same row and an integer in the same column. Thus the answer is "No".
제목: n*n의 행렬을 드리겠습니다. 행렬 안의 모든 칸이 a[i][j]를 충족시키면!=1, 그리고 a[i][j]는 그 열의 한 수와 그 줄의 한 수의 합과 같으면 Yes를 출력하고, 그렇지 않으면 No를 출력한다.
문제풀이 사고방식: 각 칸이 만족하는지 폭력적으로 판단한다
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[60][60];
int n;

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				scanf("%d", &a[i][j]);
		int flag = 1;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (a[i][j] == 1) continue;
				int flag1 = 0;
				for (int k = 1; k <= n; k++)
				{
					for(int p=1;p<=n;p++)
						if (a[i][j] == a[i][k] + a[p][j]) { flag1 = 1; break; }
					if (flag1) break;
				}
				if (!flag1) { flag = 0; break; }
			}
			if (!flag) break;
		}
		if (flag) printf("Yes
"); else printf("No
"); } return 0; }

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