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:
- g(45,3)=9g(45,3)=9 (4545 is divisible by 32=932=9 but not divisible by 33=2733=27),
- g(63,7)=7g(63,7)=7 (6363 is divisible by 71=771=7 but not divisible by 72=4972=49).
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:
- f(30,70)=g(70,2)⋅g(70,3)⋅g(70,5)=21⋅30⋅51=10f(30,70)=g(70,2)⋅g(70,3)⋅g(70,5)=21⋅30⋅51=10,
- f(525,63)=g(63,3)⋅g(63,5)⋅g(63,7)=32⋅50⋅71=63f(525,63)=g(63,3)⋅g(63,5)⋅g(63,7)=32⋅50⋅71=63.
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; }