poj 3213 PM3

题目链接：http://poj.org/problem?id=3213

解题思路：直接模拟矩阵相乘

/**************************************************************************
user_id: SCNU20102200088
problem_id: poj 3213
problem_name: PM3
**************************************************************************/

#include <algorithm>
#include <iostream>
#include <iterator>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;

//线段树
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

//手工扩展栈
#pragma comment(linker,"/STACK:102400000,102400000")

const double EPS=1e-9;
const double PI=acos(-1.0);
const double E=2.7182818284590452353602874713526;  //自然对数底数
const double R=0.5772156649015328606065120900824;  //欧拉常数:(1+1/2+...+1/n)-ln(n)

const int x4[]={-1,0,1,0};
const int y4[]={0,1,0,-1};
const int x8[]={-1,-1,0,1,1,1,0,-1};
const int y8[]={0,1,1,1,0,-1,-1,-1};

typedef long long LL;

typedef int T;
T max(T a,T b){ return a>b? a:b; }
T min(T a,T b){ return a<b? a:b; }
T gcd(T a,T b){ return b==0? a:gcd(b,a%b); }
T lcm(T a,T b){ return a/gcd(a,b)*b; }

///////////////////////////////////////////////////////////////////////////
//Add Code:
int N,P,M,A[1005][1005],B[1005][1005],C[1005][1005];
///////////////////////////////////////////////////////////////////////////

int main(){
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    ///////////////////////////////////////////////////////////////////////
    //Add Code:
    int i,j,k;
    while(scanf("%d%d%d",&N,&P,&M)!=EOF){
        for(i=1;i<=N;i++){
            for(j=1;j<=P;j++) scanf("%d",&A[i][j]);
        }
        for(i=1;i<=P;i++){
            for(j=1;j<=M;j++) scanf("%d",&B[i][j]);
        }
        for(i=1;i<=N;i++){
            for(j=1;j<=M;j++) scanf("%d",&C[i][j]);
        }
        bool flag=1;
        int row=0,col=0,ans=0;
        for(i=1;i<=N;i++){
            for(j=1;j<=M;j++){
                int temp=0;
                for(k=1;k<=P;k++) temp+=A[i][k]*B[k][j];
                if(temp!=C[i][j]){
                    flag=0;
                    row=i;
                    col=j;
                    ans=temp;
                    goto out;
                }
            }
        }
        out:;
        if(flag) printf("Yes
");
        else{
            printf("No
");
            printf("%d %d
",row,col);
            printf("%d
",ans);
        }
    }
    ///////////////////////////////////////////////////////////////////////
    return 0;
}

/**************************************************************************
Testcase:
Input:
2 3 2
1 2 -1
3 -1 0
-1 0
0 2
1 3
-2 -1
-3 -2
Output:
No
1 2
1
**************************************************************************/