P5221 Product(lcm/gcd)

最新推荐文章于 2021-05-16 15:30:35 发布

Thinking knife

最新推荐文章于 2021-05-16 15:30:35 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/zufesatoshi/article/details/90048393

本文提供了一个全面的C++算法竞赛代码模板，包括了各种常用的数据结构和算法，如数学运算、排序、搜索、字符串处理等。此外，还介绍了如何使用模板进行高效编程，以及如何在竞赛中快速定位和解决问题。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define endl '\n'
const double pi = acos(-1);
const int maxn = 1e6 + 10;
const int maxm = 5e5 + 10;
const int mod = 104857601;
bool prime[maxn];
int Prime[78500], phi[maxn], sum, cnt, n;
void ola(int n)
{
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = false;
    phi[1] = 1;
    for(int i = 2 ; i <= n ; ++ i){
        if(prime[i]){
            Prime[++cnt] = i;
            phi[i] = i - 1;
        }
        for(int j = 1 ; j <= cnt && i * Prime[j] <= n ; ++ j){
            prime[i * Prime[j]] = false;
            if(!(i %  Prime[j])){
                phi[i * Prime[j]] = phi[i] * Prime[j];
                break;
            }
            else{
                phi[i * Prime[j]] = phi[i] * phi[Prime[j]];
            }
        }
    }
}
ll f(int p){
	ll ans = 1;
	for(int i = 1 ; i <= p ; ++ i) ans = (ans * i) % mod;
	return ans % mod; 
}
ll ksm(int a, int b, int c){
    ll ans = 1, base = a % c;
    while(b){
        if(b & 1) ans = (ans * base) % c;
            base = (base * base) % c;
        b >>= 1;
    }
    return ans % c;
}
int main()
{
    ola(1e6+10);
    scanf("%d", &n);
    ll fm = 1;
    for(int i = 1 ; n / i > 1 ; ++ i){
        if(i == 1)
            fm = (fm * ksm(f(n/i)*f(n/i)%mod, phi[i], mod)) % mod;
        else
            fm = (fm * ksm(f(n/i)*f(n/i)%mod, phi[i]*2, mod)) % mod;
        //cout << ksm(f(n/i)*f(n/i)%mod, phi[i], mod) << endl;
    }
    //cout << fm << endl;
    ll ny = ksm((fm)%mod, mod - 2, mod);
    ll fz = ksm(f(n), 2*n, mod);
    //cout << fz <<endl;
    printf("%lld", (fz * ny) % mod);
    return 0;
}