D. Vasya And The Matrix
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
input
Copy
2 3
2 9
5 3 13
output
Copy
YES
3 4 5
6 7 8
input
Copy
3 3
1 7 6
2 15 12
output
Copy
NO
思路：
先说构造方式：
证明：
可以发现如果右下角的两个式子相等，就可以构造出来，即 r1^r2^r3^c1^c2^c3==0，可以发现与红色区域的取值无关。我们可以把红色区域赋为0，这样就容易构造了。
代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define MAXN 105
#define mod 19260817
int n,m,row[MAXN],col[MAXN],G[MAXN][MAXN];
int main()
{
scanf("%d%d",&n,&m);
int s=0;
for(int i=0;i<n;i++)
scanf("%d",row+i),s^=row[i];
for(int i=0;i<m;i++)
scanf("%d",col+i),s^=col[i];
if(s) puts("NO");
else
{
puts("YES");
memset(G,0,sizeof G);
for(int i=0;i<n-1;i++)
s^=row[i],G[i][m-1]=row[i];
s^=col[m-1];
G[n-1][m-1]=s;
for(int i=0;i<m-1;i++)
G[n-1][i]=col[i];
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(j) printf(" ");
printf("%d",G[i][j]);
}
printf("\n");
}
}
return 0;
}