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 print m 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 21 00 0
Output
NO
Input
2 31 1 11 1 1
Output
YES1 1 11 1 1
Input
2 30 1 01 1 1
Output
YES0 0 00 1 0第一次看题目把a，b两个数组看反了，我觉得这道题还是挺好写的，反这写就好了，只要判断a数组当行列上有1的时候对应的b数组是否为1，首先要把a数组复制出来b数组中为0是，a数组对应的行列都要为0；
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 using namespace std;
6 int a[200][200];
7 int m,n;
8 void fun(int x,int y)
9 {
10 for(int i=1;i<=m;i++)
11 a[i][y]=0;
12 for(int j=1;j<=n;j++)
13 a[x][j]=0;
14 }
15 int main()
16 {
17 int x[105],y[105];
18 int b[200][200];
19 int i,j;
20 cin>>m>>n;
21 for(i=1;i<=m;i++)
22 {
23 for(j=1;j<=n;j++)
24 a[i][j]=1;
25 }
26 memset(x,0,sizeof(x));
27 memset(y,0,sizeof(y));
28 for(i=1;i<=m;i++)
29 {
30 for(j=1;j<=n;j++)
31 {
32 cin>>b[i][j];
33 if(b[i][j]==0)
34 fun(i,j);
35 }
36 }
37 /*for(i=1;i<=m;i++)
38 {
39 for(j=1;j<=n;j++)
40 cout<<a[i][j]<<" ";
41 cout<<endl;
42 }*/
43 for(i=1;i<=m;i++)
44 {
45 for(j=1;j<=n;j++)
46 {
47 if(a[i][j]==1)
48 {
49 x[i]=1;
50 y[j]=1;
51 }
52 }
53 }
54 /*cout<<x[1]<<" "<<x[2]<<endl;
55 cout<<y[1]<<" "<<y[2]<<endl;*/
56 int flag=0;
57 for(i=1;i<=m;i++)
58 {
59 for(j=1;j<=n;j++)
60 {
61 if(x[i]==1||y[j]==1)
62 {
63 if(b[i][j]==1)
64 flag=1;
65 else
66 {
67 cout<<"NO"<<endl;
68 return 0;
69 }
70 }
71 else
72 {
73 if(b[i][j]==1)
74 {
75 cout<<"NO"<<endl;
76 return 0;
77 }
78 else
79 flag=1;
80 }
81 }
82 }
83 if(flag==1)
84 cout<<"YES"<<endl;
85 for(i=1;i<=m;i++)
86 {
87 for(j=1;j<n;j++)
88 cout<<a[i][j]<<" ";
89 cout<<a[i][n]<<endl;
90 }
91 return 0;
92 }