Steps to One

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 DEBUGfreopen("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 DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

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