PM 3

C - PM 3

USTC has recently developed the Parallel Matrix Multiplication Machine – PM³, which is used for very large matrix multiplication.

Given two matrices A and B, where A is an N × P matrix and B is a P × M matrix, PM³ can compute matrix C = AB in O(P(N + P + M)) time. However the developers of PM³ soon discovered a small problem: there is a small chance that PM³ makes a mistake, and whenever a mistake occurs, the resultant matrix C will contain exactly one incorrect element.

The developers come up with a natural remedy. After PM³ gives the matrix C, they check and correct it. They think it is a simple task, because there will be at most one incorrect element.

So you are to write a program to check and correct the result computed by PM³.

Input

The first line of the input three integers N, P and M (0 < N, P, M ≤ 1,000), which indicate the dimensions of A and B. Then follow N lines with Pintegers each, giving the elements of A in row-major order. After that the elements of B and C are given in the same manner.

Elements of A and B are bounded by 1,000 in absolute values which those of C are bounded by 2,000,000,000.

Output

If C contains no incorrect element, print “Yes”. Otherwise print “No” followed by two more lines, with two integers r and c on the first one, and another integer v on the second one, which indicates the element of C at row r, column c should be corrected to v.

两个矩阵的乘积得到一个新的矩阵，新的矩阵中可能有一个错误，如果有就输出No，并且输出错误的行和列以及正确的数应该是多少，A*B=C，那么就是A的每行乘以B的每列，【C11+C12=A11*(B11+B21)+A12*(B12+B22)】因此，先把B和C的每行之和分别算出来，用A的第i行中第j个数依次乘以B的第j行之和如果不等于C的第i行之和，则说明错误出现在这一行，然后对C中这行的每个数进行判断，找出错误的列。

#include <iostream>#include<cstdio>#include<cstring>#define NN 1005using namespace std;int a[NN][NN],b[NN][NN],c[NN][NN],fc[NN],fb[NN];int main(){    int n,p,m;    while(~scanf("%d%d%d",&n,&p,&m))    {        memset(fb,0,sizeof(fb));        memset(fc,0,sizeof(fc));        for(int i=0;i<n;i++)            for(int j=0;j<p;j++)            {                scanf("%d",&a[i][j]);            }        for(int i=0;i<p;i++)            for(int j=0;j<m;j++)            {                scanf("%d",&b[i][j]);                fb[i]+=b[i][j];            }        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)            {                scanf("%d",&c[i][j]);                fc[i]+=c[i][j];            }        int x,y,flag=1,sum,ans;        for(int i=0;i<n;i++)        {            sum=0;            for(int j=0;j<p;j++)            {                sum+=a[i][j]*fb[j];            }                if(sum!=fc[i])                {                    x=i;                    flag=0;                    break;                }        }        for(int i=0;i<m;i++)        {            sum=0;            for(int j=0;j<p;j++)            {                sum+=a[x][j]*b[j][i];            }if(sum!=c[x][i])                {                    y=i;                    ans=sum;                    break;                }        }        if(!flag)        {            printf("No\n");            printf("%d %d\n",++x,++y);            printf("%d\n",ans);        }        else printf("Yes\n");    }    return 0;}