B. OR in Matrix
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 mrows consisting of n integers representing matrix A that can produce given matrix B. If there are several solutions print any one.
Sample test(s)
input
2 21 00 0
output
NO
题意：给你一个矩阵，推原矩阵
题解：模拟就好了
//1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define meminf(a) memset(a,127,sizeof(a));
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){
if(ch==‘-‘)f=-1;ch=getchar();
}
while(ch>=‘0‘&&ch<=‘9‘){
x=x*10+ch-‘0‘;ch=getchar();
}return x*f;
}
//****************************************
#define maxn 1000+5
#define mod 1000000007
int c[maxn][maxn],b[maxn][maxn],a[maxn][maxn];
int main(){
int n=read();
int m=read();
memset(b,-1,sizeof(b));
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&a[i][j]);
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]==0){
for(int k=1;k<=m;k++){
b[i][k]=0;
}
for(int k=1;k<=n;k++){
b[k][j]=0;
}
}
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(b[i][j]==-1)b[i][j]=1;
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
int aa=0;
for(int k=1;k<=m;k++){
aa|=b[i][k];
}
for(int k=1;k<=n;k++){
aa|=b[k][j];
}
c[i][j]=aa;
}
}bool flag=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(c[i][j]!=a[i][j])flag=1;
}
}
if(!flag){
cout<<"YES"<<endl;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cout<<b[i][j]<<" ";
}
cout<<endl;
}
}
else cout<<"NO"<<endl;
return 0;
}
代码