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【YBTOJ】【Luogu】古代猪文

一天清晨 2022-02-06 阅读 39

在这里插入图片描述

题目要求 g ∑ d ∣ n C n d m o d    999911659 g^{\sum_{d|n}C_n^d} mod ~~999911659 gdnCndmod  999911659
根据拓展欧拉定理
g ∑ d ∣ n C n d    m o d    999911658 m o d    999911659 g^{\sum_{d|n}C_n^d ~~mod ~~ 999911658} mod ~~999911659 gdnCnd  mod  999911658mod  999911659
把999911658分解质因数,得到2,3,4679, 35617
分别做lucas,然后用中国剩余定理和并

c o d e code code

#include<iostream>
#include<cstdio>
#define ll long long

using namespace std;

ll P=999911658;
ll g, n, p[5]={0, 2, 3, 4679, 35617}, q[5];

ll qpow(ll x, ll t, ll mod)
{
	ll sum=1ll;
	while(t)
	{
		if(t&1)
			sum=sum*x%mod;
		x=x*x%mod;
		t>>=1;
	}
	return sum%mod;
}

ll inv(ll x, ll mod)
{
	return qpow(x, mod-2, mod);
}

ll C(ll n, ll m, ll mod)
{
	if(n<m)
		return 0;
	if(m==0)
		return 1;
	ll sum=1;
	for(ll i=n-m+1; i<=n; i++)
		sum=sum*i%mod;
	ll chu=1;
	for(ll i=1; i<=m; i++)
		chu=chu*i%mod;
	return sum*inv(chu, mod)%mod;
}

ll lucas(ll n, ll m, ll mod)
{
	if(n<m)
		return 0;
	if(m==0)
		return 1;
	return C(n%mod, m%mod, mod)*lucas(n/mod,m/mod, mod);
}

void a()
{
	for(ll i=1; i*i<=n; i++)
	{
		if(n%i==0)
		{
			for(ll j=1; j<=4; j++)
				q[j]=(q[j]+lucas(n, i, p[j]))%p[j];
			if(i*i!=n)
			{
				for(ll j=1; j<=4; j++)
					q[j]=(q[j]+lucas(n, n/i, p[j]))%p[j];
			}
		}
	}
}

ll b()
{
	ll ans=0, mo1=P;
	for(ll i=1; i<=4; i++)
	{
		ll m=mo1/p[i];
		ll t=inv(m, p[i]);
		ans=(ans+m*t*q[i])%mo1;
	}
	return ans;
}

int main()
{
	scanf("%lld%lld", &n, &g);
	g%=P+1;
	if(g==0)
	{
		printf("0");
		return 0;
	}
	a();
	ll k=(b()+P)%P;
	printf("%lld", qpow(g, k, P+1));
	return 0;
}
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