https://codeforces.com/contest/1139/problem/D
题解:莫比乌斯+无穷级数
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
ll powl(ll a, ll n, ll p) //快速幂 a^n % p
{
ll ans = 1;
while(n)
{
if(n & 1) ans = ans * a % p;
a = a * a % p;
n >>= 1;
}
return ans;
}
ll niYuan(ll a, ll b) //费马小定理求逆元
{
return powl(a, b - 2, b);
}
ll cal(ll a, ll b) //计算C(a, b)
{
return a*niYuan(b,MOD) % MOD;
}
int Miu[N] , Prime[N] , Pris ;
bool P[N] ;
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
Miu[1] = 1 ;
for(register int i = 1 ; ++i <= n ; ) {
if( not P[i] ) {
Prime[++Pris] = i ;
Miu[i] = 1 ;
}
for(register int j = 0 ; ++j <= Pris ; ) {
register int S = i * Prime[j] ;
if( S > n ) break ;
P[S] = 1 ;
if( i % Prime[j] ) Miu[S] = -Miu[i] ;
else {
Miu[S] = 0 ;
break ;
}
}
}
int Ansa = 1 , Ansb = 1 ;
for(register int i = 1 ; ++i <= n ; ) {
if( Miu[i] == 0 ) continue ;
register int Thea = n / i , Theb = n - Thea ;
if( Miu[i] == -1 ) Thea = MOD - Thea ;
Ansa = ( 1ll * Ansa * Theb + 1ll * Ansb * Thea ) % MOD;
Ansb = 1ll * Ansb * Theb % MOD ;
}
return not printf( "%lld\n" , cal(Ansa,Ansb) ) ;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}