[洛谷P4980]【模板】Pólya 定理

前言

模板题。

题目

洛谷

讲解

直接由 Pólya 定理得到下面的式子，然后暴力求就过了。(懒得对齐了，将就看吧)

\[\sum_{i=1}^{n} n^{\gcd(i,n)}=n\times ans\\ \frac{1}{n}\sum_{i=1}^nn^{\gcd(i,n)}\\ =\frac{1}{n}\sum_{d}n^{d}\sum_{i=1}^{\frac{n}{d}}[\gcd(\frac{n}{d},i)=1]\\ =\frac{1}{n}\sum_{d}n^{d}\varphi(\frac{n}{d})\\ \]

完全不能接受，凭什么暴力能过啊！！！而且线性筛预处理一点甚至还要慢一些，我不能理解。

代码

//12252024832524
#include <bits/stdc++.h>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 100005;
const int MOD = 1e9+7;
int n;

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x); if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

int qpow(int x,int y)
{
	int ret = 1;
	while(y){if(y & 1) ret = 1ll * ret * x % MOD;x = 1ll * x * x % MOD;y >>= 1;}
	return ret;
}
LL phi(int x)
{
	int ret = x;
	for(int i = 2;i*i <= x;++ i)
		if(x % i == 0)
		{
			while(x % i == 0) x /= i;
			ret = ret / i * (i-1);
		}
	if(x > 1) ret = ret / x * (x-1);
	return ret;
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	for(int T = Read(); T ;-- T)
	{
		n = Read(); LL mi = 1,ans = 0;
		for(int d = 1;d*d <= n;++ d)
		{
			mi = mi * n % MOD;
			if(n % d == 0)
			{
				ans = (ans + mi * phi(n/d)) % MOD;
				if(n/d != d) ans = (ans + qpow(n,n/d) * phi(d)) % MOD;
			}
		}
		Put(ans * qpow(n,MOD-2) % MOD,'\n');
	}
	return 0;
}