牛客多校(2020第四场)B Basic Gcd Problem（质因数分解）

题目链接：https://ac.nowcoder.com/acm/contest/5669/B

题意：

那本题只要求出n的质因子个数即可

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<vector>
using namespace std;

#define ll long long
const int N = 1e6 + 5;
const ll MOD = 1e9 +7;

int c,n;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t;
    cin>>t;
    
    while(t--) {
        scanf("%d %d",&n,&c);
        ll cnt = 1;
        
        for (int i = 2; i * i <= n; i++) {
            while (n % i == 0) {
                cnt = cnt * c % MOD;
                n /= i;
            }
        }
        if(n!=1) cnt = cnt * c % MOD;
        cout << cnt << "
";
    } 
    return 0;
}

 解法2：先将1~N的质因子个数存储起来再使用

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<vector>
using namespace std;
const int N = 1e6 + 5;
const int MOD = 1e9 +7;

long long c,n;
long long v[N], prime[N], temps[N];

void primes(int n)  {
    memset(v, 0, sizeof(v));
    memset(prime, 0, sizeof(prime));
    fill(temps, temps+N, 1); //存储质因数个数

    int m = 0; //质数数量
    for (int i = 2; i <= n; i++) {
        if (v[i] == 0)  {v[i] = i; prime[++m] = i;}

        for (int j = 1; j <= m; j++) {
            if(prime[j] > v[i] || prime[j] > n / i) break; //要赋给i * prime[j]的最小质因子为prime[j]，如果prime[j]比i的最小质因子都大，那肯定得退出循环了
            v[i*prime[j]] = prime[j];
            temps[i*prime[j]] = temps[prime[j]] + temps[i]; //递推该质因数的个数
        }
    }
}

long long fun(long long x,long long n) {
    long long res = 1;
    while (n > 0) {
        if (n & 1)  res = res * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return res;
}

int main() {
    int t;
    cin>>t;
    primes(N);    
    while(t--) {
        scanf("%lld %lld",&n,&c);
        if (n == 1)
            cout << "1
";
        else
            cout<<fun(c,temps[n]) % MOD<<"
";
    } 
    return 0;
}