【Codeforces Round #420 (Div. 2) A】Okabe and Future Gadget Laboratory

【题目链接】:http://codeforces.com/contest/821/problem/A

【题意】



给你一个n*n的数组;

然后问你,是不是每个位置(x,y);

都能找到一个同一行的元素q和同一列的元素w;

使得q+w=a[x][y]

【题解】



O(N4)模拟



【Number Of WA】



0



【反思】



不用考虑会选到a[x][y]本身.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0) typedef pair<int,int> pii;
typedef pair<LL,LL> pll; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 50; int a[N+10][N+10],n; void out(){
cout <<"No"<<endl;
exit(0);
} int main(){
//Open();
Close();
cin >> n;
rep1(i,1,n){
rep1(j,1,n){
cin >> a[i][j];
}
}
rep1(i,1,n){
rep1(j,1,n){
if (a[i][j]!=1){
int fi = 0;
rep1(ii,1,n)
rep1(jj,1,n)
if (a[ii][j]+a[i][jj]==a[i][j])
fi = 1;
if (!fi) out();
}
}
}
cout <<"Yes"<<endl;
return 0;
}
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