D. Vasya And The Matrix(Educational Codeforces Round 48)

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原文链接：http://www.cnblogs.com/huangjiaming/p/9420472.html

本文探讨了如何帮助Vasya解决数学考试中的一道难题：寻找满足特定约束条件的矩阵，包括行与列的异或值。文章提供了一种算法解决方案，通过构造一个特定的矩阵来满足题目要求。

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.

直接构造一个任意的(n-1)*(m-1)的0矩阵，最后一个值异或得到

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
using namespace std;
#define ll long long
int main(int argc, char const *argv[])
{
    int n, m;
    int r, c;
    int a[105], b[105];

    cin >> n >> m;

    r = c = 0;

    for (int i = 0; i < n; i++) {
        cin >> a[i];
        r ^= a[i];
    }
    for (int i = 0; i < m; i++) {
        cin >> b[i];
        c ^= b[i];
    }
    if (r != c) {
        puts("NO");
    } else {
        bool first;
        puts("YES");
        for (int i = 0; i < n; i++) {
            first = true;
            for (int j = 0; j < m; j++) {
                if (!first) printf(" ");
                else first = false;
                if (i != n-1 && j != m-1) {
                    printf("0");
                } else if (j == m-1 && i == n-1) {
                    r = 0;
                    for (int k = 0; k < m-1; k++) r ^= b[k];
                    r ^= a[n-1];
                    printf("%d", r);
                } else if (j == m-1){
                    printf("%d", a[i]);
                } else {
                    printf("%d", b[j]);
                }
            }
            printf("\n");
        }
    }

    return 0;
}

转载于:https://www.cnblogs.com/huangjiaming/p/9420472.html