?
2
3
4
10
100
1 1
1 1
2 1
2 2
4 6
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = (1 << 20) + 10;
int primes[N], cnt;
int minp[N];
bool st[N];//判断是否为质数
//求n以内所以质数
void get_primes(int n)
{
for (int i = 2; i <= n; i++)
{
if (!st[i]) //是质数
{
minp[i] = i;
primes[cnt++] = i;
}
for (int j = 0; primes[j] * i <= n; j++)
{
int t = primes[j] * i;//是质数的倍数
st[t] = true;//不是质数,是合数
minp[t] = primes[j];//合数t的最小质数
if (!t)break;//防止死循环
}
}
}
int main()
{
get_primes(N-1);
int num[N];//每种质因子的个数
int x;
while (scanf("%d", &x) != EOF)
{
int k=0; //质因子种类数量
int total = 0;//质因子总数
while (x > 1)
{
int p = minp[x];
num[k]=0;
while (x % p == 0)
{
x /= p;
total++;
num[k]++;
}
k++;
}
LL res = 1;
for (int i = 1; i <= total; i++)res *= i;
for (int i = 0; i < k; i++)
for (int j = 1; j <= num[i]; j++)
res /= j;
cout << total <<" "<< res << endl;
}
}