CF1228C. Primes and Multiplication（数学）

Let’s introduce some definitions that will be needed later.

Let 𝑝𝑟𝑖𝑚𝑒(𝑥) be the set of prime divisors of 𝑥. For example, 𝑝𝑟𝑖𝑚𝑒(140)={2,5,7}, 𝑝𝑟𝑖𝑚𝑒(169)={13}.

Let 𝑔(𝑥,𝑝) be the maximum possible integer 𝑝𝑘 where 𝑘 is an integer such that 𝑥 is divisible by 𝑝𝑘. For example:

𝑔(45,3)=9 (45 is divisible by 32=9 but not divisible by 33=27),
𝑔(63,7)=7 (63 is divisible by 71=7 but not divisible by 72=49).
Let 𝑓(𝑥,𝑦) be the product of 𝑔(𝑦,𝑝) for all 𝑝 in 𝑝𝑟𝑖𝑚𝑒(𝑥). For example:

𝑓(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=63.
You have integers 𝑥 and 𝑛. Calculate 𝑓(𝑥,1)⋅𝑓(𝑥,2)⋅…⋅𝑓(𝑥,𝑛)mod(109+7).

Input
The only line contains integers 𝑥 and 𝑛 (2≤𝑥≤109, 1≤𝑛≤1018) — the numbers used in formula.

Output
Print the answer.

Examples
inputCopy
10 2
outputCopy
2
inputCopy
20190929 1605
outputCopy
363165664
inputCopy
947 987654321987654321
outputCopy
593574252
Note
In the first example, 𝑓(10,1)=𝑔(1,2)⋅𝑔(1,5)=1, 𝑓(10,2)=𝑔(2,2)⋅𝑔(2,5)=2.

In the second example, actual value of formula is approximately 1.597⋅10171. Make sure you print the answer modulo (109+7).

In the third example, be careful about overflow issue.

思路：求出x的素因子，求其在1-n中所有数的贡献

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
typedef long long LL;
typedef long long ll;
const int MAXN = 150000 + 5;
const int mod = 1e9 + 7;
 
vector<LL>v;
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;
}
 
 
void primeFactor(LL n){
    LL tmp = n;
    if(n % 2 == 0) {
        v.push_back(2);
        while (n % 2 == 0) {
            n /= 2;
        }
    }
    for(LL i = 3; i * i <= tmp; i += 2){
        if(n % i == 0) {
            v.push_back(i);
        }
        while(n % i == 0){
            n /= i;
        }
    }
    if(n > 2)
        v.push_back(n);
}
int main()
{
    LL x,n;
    cin >> x;
    cin >> n;
    primeFactor(x);
    LL ans = 1;
    for(int i = 0; i < v.size(); i++) {
        LL tt = 0,nn = n;
        while(nn > 0) {
            nn /= v[i];
            tt += nn;
        }
        ans = (ans % mod * qpow(v[i],tt)) % mod;
    }
 
    cout << ans << endl;
 
}