【COderForces】OR in Matrix

OR in Matrix
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

Description

Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if
either or both of the logical values is set to 1, otherwise it is 0. We can define logical OR of
three or more logical values in the same manner:


 where 

 is
equal to 1 if some ai = 1, otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns. The
rows are numbered from 1 to m, columns are numbered from 1 to n.
Element at row i (1 ≤ i ≤ m) and column j (1 ≤ j ≤ n)
is denoted as Aij. All elements of A are either 0 or 1. 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)

Nam gives you matrix B and challenges you to guess matrix A. Although Nam is smart, he could probably make a mistake while calculating
matrix B, since size of A can be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100),
number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated by spaces describing rows of matrix B (each
element of B is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B, otherwise print "YES". If the first
line is "YES", then also printm rows consisting of n integers representing matrix A that
can produce given matrix B. If there are several solutions print any one.

Sample Input

Input

2 2
1 0
0 0

Output

NO

Input

2 3
1 1 1
1 1 1

Output

YES
1 1 1
1 1 1

Input

2 3
0 1 0
1 1 1

Output

YES
0 0 0
0 1 0

矩阵的计算；

代码：

#include<stdio.h>
int a[105][105],b[105][105];
int main()
{
int n,m;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&b[i][j]);
a[i][j]=1;
}
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(!b[i][j])
{
for(int k=1;k<=m;k++)
a[k][j]=0;
for(int m1=1;m1<=n;m1++)
a[i][m1]=0;
}
}
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(b[i][j])
{
int f=0;
for(int k=1;k<=n;k++)
{
f|=a[i][k];
}
for(int k=1;k<=m;k++)
{
f|=a[k][j];
}
if(!f)
{
printf("NO\n");
return 0;
}
}
}
}
printf("YES\n");
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(j==1)
printf("%d",a[i][j]);
else
printf(" %d",a[i][j]);
}
printf("\n");
}
}
return 0 ;
}