Day 9 C. Primes and Multiplication_primes and xor? nonsense-优快云博客

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这道数学问题涉及定义素数因子集合、最大整数幂以及两个数的乘积函数。给定整数x和n，求f(x,1) * f(x,2) * ... * f(x,n)对10^9 + 7取模的结果。需要注意大数处理和溢出问题。" 105727557,8320156,CSS实现下拉及多级弹出菜单技巧解析,"['前端开发', 'CSS', '网页布局']

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Problem:
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

input
10 2
output
2

input
20190929 1605
output
363165664

input
947 987654321987654321
output
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.

题目大致意思…数学题具体解释看代码注释

#include<iostream>
#include<cmath>
#include<algorithm>
#define ll long long 
using namespace std;
const int mod = 1e9 + 7;
ll p[200005];
ll c[200005];
ll m;
//题目中g函数g(x,p)表明p在x中最多能被整除的次数 
//f函数(x,y)表明将x分解质因数后 g(y,不重复的质因数1)*g(y,不重复的质因数2)*...*g(y,不重复的质因数n)
void f(ll  n)//记录一个质数在这个n里面出现的次数 相当于g函数中的p
{
    m = 0;
    ll len = sqrt(n);
    for (ll i = 2; i <= len; i++)
    {
        if (n % i == 0)
        {
            p[++m] = i;
            c[m] = 0;
            while (n % i == 0)
            {
                n /= i;
                c[m]++;
            }
        }
    }
    if (n > 1)
    {
        p[++m] = n;
        c[m] = 1;
    }
}
ll pow_mod(ll  a, ll b)//快速幂取模
{
    ll res = 1;
    for (; b; b >>= 1)
    {
        if (b & 1)
        {
            res = res * a % mod;
        }
        a = a * a % mod;
    }
    return res;
}
ll ff(ll a, ll b)//计算f函数
{
    ll res = 0;
    for (ll i = b; i <= a;i)
    {
        res += (1ll * a / i);
        a /= i;
    }
    return pow_mod(b, res);
}
int main()
{
    ll x;
    ll n;
    cin >> x >> n;
    f(x);
    ll ans = 1;//f(x,1)*f(x,2)*...*f(x,n)的结果
    for (int i = 1; i <= m; i++)
    {
        ll t = 1;
        t = ff(n, p[i]);//f(x,i)的结果
        ans = ans * t % mod;
    }
    cout << ans << endl;
    return 0;
}