超超的中等意思
http://120.78.162.102/problem.php?cid=1432&pid=11
http://120.78.162.102/problem.php?id=6249
題解：
參考文章：
https://blog.csdn.net/arrowlll/article/details/52629448?
https://blog.csdn.net/weixin_43272781/article/details/85311412
快速冪+快速組合數
/* *@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=50000+10; const lll MOD=(1ll<<61)-1; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; lll Jc[N]; void print(lll x){if(x==0)return;print(x/10);int tmp=x%10;printf("%d",tmp); } void calJc(){ //求maxn以內的數的階乘Jc[0] = Jc[1] = 1;for(lll i = 2; i < N; i++){Jc[i] = Jc[i - 1] * i %MOD;} } lll PowerMod(lll a, lll b,lll c){lll ans = 1;a = a % c;while(b>0){if(b % 2 == 1)ans = (ans * a) % c;b >>= 1;a = (a * a) % c;}return ans; } lll niYuan(lll a, lll b){ //費馬小定理求逆元return PowerMod(a, b - 2, b); }lll C(ll a, ll b){ //計算C(a, b)return Jc[a] * niYuan(Jc[b]*Jc[a - b]% MOD, MOD) % MOD; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifcalJc();int k,p,q,a,b;while(scanf("%d%d%d%d%d",&p,&q,&k,&a,&b)!=EOF){print((C(k, a)*(PowerMod(p,a,MOD)*PowerMod(q,b,MOD)%MOD))%MOD);printf("\n");}//cout << "Hello world!" << endl;return 0; }?
總結
