https://codeforces.com/contest/1139/problem/D

題解：莫比烏斯+無窮級數

/* *@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) //快速冪 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) //費馬小定理求逆元 {return powl(a, b - 2, b); }ll cal(ll a, ll b) //計算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; }

?

總結

以上是生活随笔為你收集整理的Steps to One的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。