Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造

2019-08-26

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造

技术图片


【Problem Description】

? 给你一个\(n\),构造一个\(n\times n\)的矩阵,使其满足任意一行,或一列的异或值相同。保证\(n\)能被\(4\)整除。

【Solution】

? 可以发现,从\(0\)开始,每\(4\)个连续数的异或值为\(0\),所以可以很容易使得每一行的异或值等于\(0\)

? 列向可以发现,从\(0\)开始,公差为\(4\)的,每\(4\)个数的异或值也为\(0\)。即\(0\oplus4\oplus8\oplus12=0\)。所以满足条件的矩阵形如:
\[ \left[\begin{matrix} 0&1&2&3&32&33&34&35&\4&5&6&7&36&37&38&39&\8&9&10&11&40&41&42&43&\12&13&14&15&44&45&46&47&\16&17&18&19&48&49&50&51&\20&21&22&23&52&53&54&55&\24&25&26&27&56&57&58&59&\28&29&30&31&60&61&62&63&\\end{matrix}\right] \]


【Code】

/*
 * @Author: Simon 
 * @Date: 2019-08-26 18:38:05 
 * @Last Modified by: Simon
 * @Last Modified time: 2019-08-26 19:02:53
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 2005
int a[maxn][maxn];
Int main(){
#ifndef ONLINE_JUDGE
    //freopen("input.in","r",stdin);
    //freopen("output.out","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;cin>>n;
    for(int i=0;i<n;i++) a[0][i]=i/4*n*4+i%4;
    for(int i=1;i<n;i++){
        for(int j=0;j<n;j++){
            a[i][j]=a[i-1][j]+4;
        }
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cout<<a[i][j]<<' ';
        }
        cout<<endl;
    }
#ifndef ONLINE_JUDGE
    system("pause");
#endif
    return 0;
}

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造

Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造

原文地址:https://www.cnblogs.com/--Simon/p/11414317.html