AtCoder Beginner Contest 110 D

D - Factorization

思路：把相同的质因子看成相同的小球，求把这些小球放进n个盒子里的方案数。

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL unsigned long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 10;
const int MOD = 1e9 + 7;
int fac[N], inv[N];
int cnt[N];
LL q_pow(LL n, LL k) {
    LL ans = 1;
    while(k) {
        if(k&1) ans = (ans * n) % MOD;
        n = (n * n) % MOD;
        k >>= 1;
    }
    return ans;
}
void init() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) {
        fac[i] = (1LL * fac[i-1] * i) % MOD; 
    }
    inv[N-1] = q_pow(fac[N-1], MOD-2) % MOD;
    for (int i = N-2; i >= 0; i--) inv[i] = (1LL * inv[i+1] * (i+1)) % MOD;
}
LL C(int n, int m) {
    return ((1LL * fac[n] * inv[m]) % MOD * inv[n-m]) % MOD;
}
int main() {
    int n, m, up = 0;
    init();
    scanf("%d %d", &n, &m);
    for (int i = 2; i*i <= m; i++) {
        if(m % i == 0) {
            int tmp = 0;
            while(m % i == 0) m /= i, tmp++;
            cnt[++up] = tmp;
        }
    }
    if(m > 1) cnt[++up] = 1;
    LL ans = 1;
    for (int i = 1; i <= up; i++) ans = (ans * C(cnt[i]+n-1, n-1)) % MOD;
    printf("%lld
", ans);
    return 0;
}