[AGC027D] Modulo Matrix

编程入门 行业动态 更新时间:2024-10-24 12:33:56

[AGC027D] <a href=https://www.elefans.com/category/jswz/34/1660517.html style=Modulo Matrix"/>

[AGC027D] Modulo Matrix

看兔爷的吧:
这谁想得出来啊
ATC NB!!
code:


#include<bits/stdc++.h>
#define ll long long
#define N 1000005
using namespace std;
int vis[N], prime[N], n, sz;
ll a[1005][1005];
void get_prime() {for(int i = 2; i < N; i ++) {if(!vis[i]) prime[++ sz] = i;for(int j = 1; j <= sz && prime[j] * i < N; j ++) {vis[i * prime[j]] = 1;if(i % prime[j] == 0) break;}}
}
ll gcd(ll x, ll y) {if(!x || !y) return x + y;return y? gcd(y, x % y) : x;
}
ll lcm(ll x, ll y) {if(!x || !y) return x + y;return x / gcd(x, y) * y;
}
int main() {get_prime();scanf("%d", &n);if(n == 2) {printf("4 7\n23 10"); return 0;}for(int i = 1; i <= n; i ++)for(int j = 1; j <= n; j ++)if(!((i + j) & 1)) a[i][j] = (ll)prime[(i + j) / 2] * prime[(i + n - j + 1) / 2 + n];for(int i = 1; i <= n; i ++)for(int j = 1; j <= n; j ++)if((i + j) & 1) a[i][j] = lcm(lcm(a[i - 1][j], a[i + 1][j]), lcm(a[i][j - 1], a[i][j + 1])) + 1;for(int i = 1; i <= n; i ++) {for(int j = 1; j <= n; j ++) printf("%lld ", a[i][j]); printf("\n");}return 0;
}

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[AGC027D] Modulo Matrix

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