CodeForces - 1228C (prime factorisation + Contribution Method)

The meaning of problems

https://vjudge.net/problem/CodeForces-1228C

First, start with some definitions related to:

Defined Prime (x) denotes the set of prime factors of x. For example, P R & lt I m E ( . 1 . 4 0 ) = { 2 , . 5 , . 7 }, P R & lt I m E ( . 1 . 6 . 9 ) = { . 1 . 3 }.

Definition of G ( x , P ) indicating the presence of a maximum K ∈ N *, x may be such that p ^ k divisible , then g (x, p) = p ^ k . for example:

• G (45, 3) = 9  ( 45 may be 3 ^ 2 = 9 divisible but can not be 3 ^ 3 = 27 divisible)
• G (63 is, 7) = 7  (63 is may be 7 ^ 7 = 1 divisible but can not be 7 ^ 2 = 49 divisible)

Definition of f (x, y) represents all G ( Y , P ) ( P ∈ P R & lt I m E ( X ) product) is, for example:  

• f(30, 70) = g(70,2)·g(70,3)·g(70, 5) = 2^1·3^0·5^1 = 10
  
• f(525,63) = g(63,3)·g(63,5)·g(63,7) = 3^2·5^0·7^1 = 63
  

Now given two integers x and n, the Calculate F ( x , . 1 ) ⋅ F ( x , 2 ) ... F ( x , n- ) m O D ( . 1 0 ^ . 9 + . 7 values).  

Thinking

First count x = 10, n = 10 in the case of

f(10,1)=1　　f(10,2)=g(2,2)=2　　f(10,3)=1　　f(10,4)=g(4,2)=4　　f(10,5)=g(5,5)=5

f(10,6)=g(6,2)=2　　f(10,7)=1　　f(10,8)=g(8,2)=8　　f(10,9)=1　　f(10,10)=g(10,5)=10

Easy to find, for the prime factors of 10 2,4,6,8,10 2,5,2 have emerged once again appeared in the 4,8, 8 appeared again. So for prime factor i, its contribution is x ^ (n / x) * x ^ (n / x / x) * x ^ (n / x / x / x) * ......

So the prime factorization of x (x can be decomposed to the root), then the contribution of each calculated quality factor.

Code

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
const int N=200005;
const int mod=1e9+7;
const double eps=1e-8;
const double PI = acos(-1.0);
#define lowbit(x) (x&(-x))
ll qpow(ll a,ll b)
{
    ll res=1;
    while(b)
    {
        if(b&1) res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
int main()
{
    std::ios::sync_with_stdio(false);
    ll x,n;
    cin>>x>>n;
    ll xx=x,gx=sqrt(x);
    ll ans=1;
    for(ll i=2; i<=gx; i++)
    {
        int f=0;
        if(xx==1)
            break;
        while(xx%i==0&&xx!=1)
        {
            xx/=i;
            f=1;
        }
        if(f)
        {
            ll nn=n;
            while(nn)
            {
                nn/=i;
                ans=ans*qpow(i,nn)%mod;
            }
        }
    }
    if(xx>1)
    {
        ll nn=n,i=xx;
        while(nn)
        {
            nn/=i;
            ans=ans*qpow(i,nn)%mod;
        }
    }
    cout<<ans<<endl;
    return 0;
}