[CodeForces] 1016D Vasya And The Matrix

最新推荐文章于 2021-07-10 22:41:20 发布

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最新推荐文章于 2021-07-10 22:41:20 发布

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原文链接：http://www.cnblogs.com/floatiy/p/9482959.html

本文介绍了一种矩阵构造题的解决方法，对于给定的行和列的异或和，构造一个满足条件的矩阵，或者判断无法构造。通过特定的填充策略完成矩阵构造，并给出了具体的代码实现。

题目描述

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 $a_{1},a_{2},...,a_{n} a1,a2,...,an denotes the xor of elements in rows with indices 1 1 1 , 2 2 2 , ..., n n n , respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1,b2,...,bm b_{1},b_{2},...,b_{m} b1,b2,...,bm denotes the xor of elements in columns with indices 1 1 1 , 2 2 2 , ..., m m m , respectively.$

Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.

翻译：已知一个n×m的矩阵，每行每列元素的异或和，请构造一个满足要求的矩阵。若不存在，输出"NO"，否则输出"YES"和矩阵。

输入输出格式

输入格式：

The first line contains two numbers

The second line contains $a_{1},a_{2},...,a_{n} (0<=a_{i}<=10^{9}) $ , where ai a_{i} ai is the xor of all elements in row i i i .$

The third line contains $b_{1},b_{2},...,b_{m} (0<=b_{i}<=10^{9}) $ , where bi b_{i} bi is the xor of all elements in column i i i .$

输出格式：

If there is no matrix satisfying the given constraints in the first line, output "NO".

Otherwise, on the first line output "YES", and then $c_{i1},c_{i2},...\ ,c_{im} (0<=c_{ij}<=2·10^{9}) $ — the description of the matrix.$

If there are several suitable matrices, it is allowed to print any of them.

输入输出样例

输入样例#1：

2 3
2 9
5 3 13

输出样例#1：

YES
3 4 5
6 7 8

输入样例#2：

3 3
1 7 6
2 15 12

输出样例#2：NO

题目解析

显然构造题啊，因为0 ^ n = n，所以手玩样例就可以发现我们只要最后一行填a[]，最后一列填b[]，其他全填0就好了。

这道题的难点在于处理角上的那个数（即a，b的交点），这个点暴力算出来就好了（b的异或和）。

至于无解的情况，我们用a的异或和和 b的异或和向比较，如果不相等就无解了。

证明很简单，只要时刻记着0 ^ n = n就可以了。

Code

#include<iostream>
#include<cstdio>
using namespace std;

const int MAXN = 100 + 5;

int n,m;
int a[MAXN],b[MAXN];
int ans1,ans2;

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++) {
        scanf("%d",&a[i]);
        ans1 ^= a[i];
    }
    for(int i = 1;i <= m;i++) {
        scanf("%d",&b[i]);
        ans2 ^= b[i];
    }
    if(ans1 ^ ans2) {
        printf("NO\n");
        return 0;
    } else printf("YES\n");
    for(int i = 1;i < n;i++) {
        for(int j = 1;j < m;j++) printf("0 ");
        printf("%d\n",a[i]);
    }
    for(int i = 1;i < m;i++) {
        printf("%d ",b[i]);
    }
    int ans = b[m];
    for(int i = 1;i < n;i++) ans ^= a[i];
    printf("%d",ans);
    return 0;
}