3213.txt

来自「北大ACM题目例程 详细的解答过程 程序实现 算法分析」· 文本 代码 · 共 74 行

Source

Problem Id:3213  User Id:fzk 
Memory:8684K  Time:795MS
Language:C++  Result:Accepted

Source 

#include <stdio.h>
#include <memory.h>

int a[1000][1000], sa[1000];
int b[1000][1000], sb[1000];
int c[1000][1000], sca[1000], scb[1000];
int main( ) {
	int n, p, m, i, j, k, x, y;
	scanf( "%d%d%d", &n, &p, &m );
	
	memset( sa, 0, sizeof sa );
	memset( sb, 0 ,sizeof sb );	
	
	for( i=0; i<n; i++ )
	for( j=0; j<p; j++ ) {
		scanf( "%d", &a[i][j] );
		sa[j] += a[i][j];
	}

	for( i=0; i<p; i++ )
	for( j=0 ;j<m; j++ ) {
		scanf( "%d", &b[i][j] );
		sb[i] += b[i][j];
	}

	for( i=0; i<n; i++ )
	for( j=0; j<m; j++ ) {
		scanf( "%d", &c[i][j] );
		sca[i] += c[i][j];
		scb[j] += c[i][j];
	}
	
	int sum = 0;
	for( i=0; i<n; i++ ) {
		sum = 0;
		for( j=0; j<p; j++ )
			sum += a[i][j] * sb[j];
		if( sum != sca[i] )
			break;
	}
	x = i;

	for( i=0; i<m; i++ ) {
		sum = 0;
		for( j=0; j<p; j++ )
			sum += sa[j] * b[j][i];
		if( sum != scb[i] )
			break;
	}
	y = i;

	if( x < n ) {
		printf( "No\n" );
		sum = 0;
		for( k=0; k<p; k++ )
			sum += a[x][k]*b[k][y];
		printf( "%d %d\n%d\n", x+1, y+1, sum );
	}
	else
		printf( "Yes\n" );

	return 0;
}