Educational Codeforces Round 48 (Rated for Div. 2) D. Vasya And The Matrix(构造

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本文介绍了一个关于构造特定矩阵的问题，该矩阵的每行和每列元素的异或值已知，需要找到符合条件的矩阵或者说明不存在这样的矩阵。文章提供了一种解决方案，并附带了C++代码实现。

D. Vasya And The Matrix
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
inputCopy
2 3
2 9
5 3 13
outputCopy
YES
3 4 5
6 7 8
inputCopy
3 3
1 7 6
2 15 12
outputCopy
NO

解析：直接构造一个任意的(n-1)*(m-1)的矩阵，然后通过异或得到最后一行和最后一列的值即可

#include<bits/stdc++.h>
using namespace  std;
#define ll long long
#define pb push_back
#define inf 200000099999999
#define mod 998244353
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
const int N=1e2+10;
int ar[N],ap[N];
int l[N][32];
int h[N][32];
int an[N][N];
int main()
{
    ios::sync_with_stdio(0);
    int n,m;
    cin>>n>>m;
    int mx=0,sum1=0,sum2=0;
    for(int i=0;i<n;i++)
    {
        cin>>ar[i];
        if(i==0) sum1=ar[i];
        else
            sum1=sum1^ar[i];
    }
    for(int i=0;i<m;i++)
    {
        cin>>ap[i];
        if(i==0) sum2=ap[i];
        else
            sum2=sum2^ap[i];
    }
    if(sum1!=sum2)//所有值的异或不等于所有值的异或
    {
        cout<<"NO"<<endl;
        return 0;
    }
    cout<<"YES"<<endl;
    for(int i=0;i<n-1;i++)
    {
        for(int j=0;j<m-1;j++)
        {
            an[i][j]=1;
        }
    }
    for(int i=0;i<n;i++)
    {
        int a=an[i][0];
        for(int j=1;j<m-1;j++)
        a=a^an[i][j];
        an[i][m-1]=a^ar[i];
    }
    for(int j=0;j<m;j++)
    {
        int a=an[0][j];
        for(int i=1;i<n-1;i++)
        a=a^an[i][j];
        an[n-1][j]=a^ap[j];
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
            cout<<an[i][j]<<' ';
        cout<<endl;
    }
    return 0;
}