POJ 3318 Matrix Multiplication 输入外挂z

2
1 0
2 3
5 1
0 8
5 1
10 26

YES

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 505
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
using namespace std;
int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];
inline int Scan(){
int d=0;
char c,t=0;
while((c=getchar())==' '||c=='\n');
if(c=='-')t=1;
else d=c-'0';
while((c=getchar())>='0'&&c<='9')
d=d*10+c-'0';
if(t)return -d;
else return d;
}
int main(){
int n,t;n=Scan();
FOR(i,1,n)FOR(j,1,n)a[i][j]=Scan();
FOR(i,1,n)FOR(j,1,n)b[i][j]=Scan();
FOR(i,1,n)FOR(j,1,n)c[i][j]=Scan();
FOR(i,1,n)FOR(j,1,n){
t=0;
FOR(k,1,n)t+=a[i][k]*b[k][j];
if(t!=c[i][j]){puts("NO\n");return 0;}
}
puts("YES\n");
return 0;
}
/*
2
1 0
2 3
5 1
0 8
5 1
10 26*/