标签：board second overflow header include amp ini examples turn
Let‘s introduce some definitions that will be needed later.
Let prime(x)prime(x) be the set of prime divisors of xx. For example, prime(140)={2,5,7}prime(140)={2,5,7}, prime(169)={13}prime(169)={13}.
Let g(x,p)g(x,p) be the maximum possible integer pkpk where kk is an integer such that xx is divisible by pkpk. For example:
Let f(x,y)f(x,y) be the product of g(y,p)g(y,p) for all pp in prime(x)prime(x). For example:
You have integers xx and nn. Calculate f(x,1)⋅f(x,2)⋅…⋅f(x,n)mod(109+7)f(x,1)⋅f(x,2)⋅…⋅f(x,n)mod(109+7).
The only line contains integers xx and nn (2≤x≤1092≤x≤109, 1≤n≤10181≤n≤1018) — the numbers used in formula.
Print the answer.
10 2
2
20190929 1605
363165664
947 987654321987654321
593574252
In the first example, f(10,1)=g(1,2)⋅g(1,5)=1f(10,1)=g(1,2)⋅g(1,5)=1, f(10,2)=g(2,2)⋅g(2,5)=2f(10,2)=g(2,2)⋅g(2,5)=2.
In the second example, actual value of formula is approximately 1.597⋅101711.597⋅10171. Make sure you print the answer modulo (109+7)(109+7).
In the third example, be careful about overflow issue.
#include<iostream> #include<stdio.h> #include<algorithm> #include<math.h> using namespace std; typedef unsigned long long ll; ll fac[10050], num;//素因数,素因数的个数 const ll mod=1e9+7; ll pow_mod(ll a, ll n, ll m) { if(n == 0) return 1; ll x = pow_mod(a, n/2, m); ll ans = (ll)x * x % m; if(n % 2 == 1) ans = ans *a % m; return (ll)ans; } void init(ll n) {//唯一分解定理 num = 0; ll cpy = n; ll m = (int)sqrt(n + 0.5); for (int i = 2; i <= m; ++i) { if (cpy % i == 0) { fac[num++] = i; while (cpy % i == 0) cpy /= i; } } if (cpy > 1) fac[num++] = cpy; } int main(){ ll x,n; cin>>x>>n; init(x); ll ans=1; for(ll i=0;i<num;i++){ for(ll cur=fac[i];;cur*=fac[i]){ ans=ans*pow_mod(fac[i],n/cur,mod)%mod; if(cur>n/fac[i]) break; } } cout<<ans%mod<<endl; }
C. Primes and Multiplication（数学）（防止爆精度）
标签：board second overflow header include amp ini examples turn
原文地址：https://www.cnblogs.com/ellery/p/11628053.html
