小a与黄金街道
https://ac.nowcoder.com/acm/contest/317/D
題解:歐拉函數+快速冪
參考文章:
https://blog.csdn.net/weixin_43272781/article/details/85558253
https://blog.csdn.net/weixin_43272781/article/details/85058595
/* *@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=300000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,q; int ans,cnt,temp,num; ll a,b; char str; ll euler(ll n){ //返回euler(n)ll res=n,a=n;for(int i=2;i*i<=a;i++){if(a%i==0){res=res/i*(i-1);//先進行除法是為了防止中間數據的溢出while(a%i==0) a/=i;}}if(a>1) res=res/a*(a-1);return res; } ll PowerMod(ll a, ll b, ll c){ll ans = 1;a = a % c;while(b>0){if(b % 2 == 1)ans = (ans * a) % c;b >>= 1;a = (a * a) % c;}return ans; }int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifscanf("%lld%lld%lld%lld",&n,&k,&a,&b);int phi=euler(n);//cout<<phi<<endl;cout<<((a+b)*PowerMod(k,phi/2*n,MOD))%MOD<<endl;//cout << "Hello world!" << endl;return 0; }?
總結