light oj 1236 【大数分解】

给定一个大数，分解质因数，每个质因子的个数为e1,e2,e3,……em，

则结果为((1+2*e1)*(1+2*e2)……(1+2*em)+1)/2.

//light oj 1236 大数分解素因子

#include <stdio.h>  
#include <iostream>  
#include <string.h>  
#include <algorithm>  
#include <math.h>  
#include <ctype.h>  
#include <time.h>  
#include <queue>  
#include <iterator>  

const int MAXN = 10000200;

bool com[MAXN];
int primes;
long long prime[MAXN/10];
long long ans, t, n;

void init(int n)
{
	primes = 0;
	memset(com, false, sizeof(com));
	com[0] = com[1] = true;
	for (int i = 2; i <= n; ++i)
	{
		if (!com[i])
		{
			prime[++primes] = i;
		}
		for (int j = 1; j <= primes && i*prime[j] <= n; ++j)
		{
			com[i*prime[j]] = true;
			if (!(i % prime[j]))
				break;
		}
	}
}

long long solve(long long n)//大数分解
{
	ans = 1;
	for (int i = 1; i<=primes && prime[i] * prime[i] <= n; i++)
	{
		if (n % prime[i] == 0)
		{
			t = 1;
			n /= prime[i];
			while (n%prime[i] == 0)
			{
				t++;
				n /= prime[i];
			}
			ans *= (1 + 2 * t);
		}
	}
	if (n > 1)
		ans *= 3;
	return (1+ans) / 2;
}

int main()
{ 
	init(10000000);
	int tt, cases = 1;
	scanf("%d",&tt);
	while (tt--)
	{
		scanf("%lld",&n);
		long long res = solve(n);
		printf("Case %d: %lld\n",cases++,res);
	}
	return 0;
}