Codeforces821A Okabe and Future Gadget Laboratory 2017-06-28 14:55 80人阅读 评论(0) 收藏
2 seconds
256 megabytes
standard input
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!
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).
Print "Yes" if the given lab is good and "No"
otherwise.
You can output each letter in upper or lower case.
3
1 1 2
2 3 1
6 4 1
Yes
3
1 5 2
1 1 1
1 2 3
No
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".
———————————————————————————————————
题目的意思是给出一个矩阵,问是否矩阵每一个除了1的数都可以由它所在行和所在列
的两个数相加得到
思路:暴力
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath> using namespace std; #define LL long long
const int inf=0x7fffffff; int n;
int mp[100][100]; bool ok(int x,int y)
{
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
if(mp[x][i]+mp[j][y]==mp[x][y])
return 1; return 0;
}
int main()
{
scanf("%d",&n);
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
scanf("%d",&mp[i][j]);
int flag=0;
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{ if(mp[i][j]!=1)
if(!ok(i,j))
{
flag=1;
break;
}
}
if(flag)
break;
}
if(flag)
printf("No\n");
else
printf("Yes\n");
return 0;
}
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