codeforces1485D.Multiples and Power Differences(思维)

题意:构造一个矩阵b,使得对于每个i和j,b[i][j]都是a[i][j]的倍数且矩阵b中相邻两数差值为某正整数的4次方。

思路:首先想到用a中所有数的lcm,进一步保证相邻两数不同,每隔一个数换成 lcm + a矩阵中该位置的数的四次方

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;
const int N = 520;

int a[N][N];

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b) {
    return a / gcd(a, b) * b;
}

int main() {
    int n, m, lcmm = 1;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            scanf("%d", &a[i][j]);
            lcmm = lcm(lcmm, a[i][j]);
        }
    }
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            if(j > 1) printf(" ");
            if((i + j) & 1) printf("%d", lcmm);
            else printf("%d", lcmm + a[i][j] * a[i][j] * a[i][j] * a[i][j]);
        }
        printf("\n");
    }
    return 0;
}

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转载自blog.csdn.net/weixin_43871207/article/details/113811294