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

告诉你每一行、每一列的异或和。
让你求出一个符合要求的原矩阵。

【题解】

显然应该有
a1^a2^....^an = b1^b2^....^bn
也即两边同时异或一下a1^b1
b1^a2^a3...^an =a1^b2^...^bn
我们设x=b1^a2^a3...^an =a1^b2^...^bn
然后我们就能得到一个符合要求的矩阵了
![](https://images2018.cnblogs.com/blog/1251265/201808/1251265-20180804114209048-744453766.png)
会发现第一列的异或和就是b1.第一行的异或和就是a1
因为x同时可以写成上面的那两种形式
第2到第n行以及第2到第m列显然也是满足要求的,下面全是0了

【代码】

#include <bits/stdc++.h>

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define all(x) x.begin(),x.end()

#define pb push_back

#define lson l,mid,rt<<1

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%lld",&x)

#define res(x) scanf("%s",x)

#define rson mid+1,r,rt<<1|1

using namespace std;

const double pi = acos(-1);

const int dx[4] = {0,0,1,-1};

const int dy[4] = {1,-1,0,0};

const int N =1000000;

LL m,n,arr[N+10],brr[N+10],ans[1000][1000];

int main()

{

    	#ifdef LOCAL_DEFINE

	    freopen("rush_in.txt", "r", stdin);

	#endif

    ios::sync_with_stdio(0),cin.tie(0);

    cin >> n>> m;

    int a=0,b=0;

    for (int i = 0; i < n; i++)

    {

        cin >>arr[i];

        if(i!=0) a=a^arr[i];

    }

    for (int i = 0; i < m; i++)

    {

        cin >>brr[i];

        if(i!=0) b=b^brr[i];

    }

    a=a^brr[0];

    b=b^arr[0];

    if(a!=b){

        cout << "NO" <<endl;

        return 0;

    }

    rep1(i,0,n-1){

        rep1(j,0,m-1){

            if(i==0&&j==0) ans[i][j]=a;

            else if(i==0)

            {

                ans[i][j]=brr[j];

            }

            else if(j==0)

            {

                ans[i][j]=arr[i];

            }

        }

    }

    cout << "YES" <<endl;

    rep1(i,0,n-1){

        rep1(j,0,m-1){

            if(j!=0) cout << " ";

            cout <<ans[i][j] ;

        }

        cout << endl;

    }

    return 0;

}

巴特西

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

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