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, ..., bmdenotes 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
题意：构造一个 n 行 m 列，i 行的异或和 为a[i]，j 列的异或和 为b[j]。
题解：对于这样一个矩阵，我们只需要考虑最后一行和最后一列即可，其他都用0填充。然后，最后一行和最后一列也只需要考虑第n行第m个即可。对于样例 行：2 ,9 列：5,3,13 .我们只需要构造出
0，0， 2
5，3，15
这样一个矩阵即可，只需要考虑最后一个位置的数字即可，令最后一个数字为x的话，有
2^x=13， 5^3^x=9。很容易就可得出，x=13^2=9^5^3=15.
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#include<set>
#include<map>
#include<math.h>
#include<vector>
#include<bitset>
#include<iostream>
using namespace std;
int a[105],b[105],s[105][105]={0};
int main(){
int n,m;
scanf("%d %d",&n,&m);
int ans1=1;
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
s[i][m-1]=a[i];
if(i!=n-1)
ans1^=a[i];
}
ans1^=1;
int ans2=1;
for(int i=0;i<m;i++){
scanf("%d",&b[i]);
s[n-1][i]=b[i];
if(i!=m-1)
ans2^=b[i];
}
ans2^=1;
ans1^=b[m-1];
ans2^=a[n-1];
if(ans1!=ans2){
printf("NO\n");
return 0;
}
s[n-1][m-1]=ans1;
printf("YES\n");
for(int i=0;i<n;i++){
for(int j=0;j<m;j++)
printf("%d ",s[i][j]);
printf("\n");
}
return 0;
}