codeforces 486B OR in Matrix

From matrix A, Nam creates another matrix
B of the same size using formula:


.

(Bij is
OR of all elements in row
i and column j of matrix
A)

先把矩阵A全部置为1，再把B矩阵中为0的位置，对应的A中位置的行和列置为0.就得到了原矩阵A；

/********************
 * Author:fisty
 * Data:2014-11-15
 * cf468B
 * *******************/

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAX_N 110
int a[MAX_N][MAX_N], b[MAX_N][MAX_N];
int main(){
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = 1;i <= n; i++){
                for(int j = 1; j <= m; j++){
                        scanf("%d", &b[i][j]);
                        a[i][j] = 1;
                }
        }
        for(int i = 1; i <= n; i++){
                for(int j = 1; j <= m; j++){
                        if(!b[i][j]){
                                for(int k = 1; k <= m; k++)
                                        a[i][k] = 0;
                                for(int k = 1; k <= n; k++)
                                        a[k][j] = 0;
                        }
                }
        }
        for(int i = 1;i <= n; i++){
                for(int j = 1; j <= m; j++){
                        if(b[i][j]){
                                int f = 0;
                                for(int k = 1;k <= m; k++)
                                        f |= a[i][k];
                                for(int k = 1; k <= n; k++)
                                        f |= a[k][j];
                                if(!f){
                                        printf("NO\n");
                                        return 0;
                                }
                        }
                }
        
        }
        printf("YES\n");       
        for(int i = 1; i <= n; i++){
                for(int j = 1; j <= m; j++){
                        printf("%d ",a[i][j]);
                        
                }
                printf("\n");
        }

        return 0;

}