light oj 1236 【大数分解】

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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+2e1)(1+2e2)……(1+2em)+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;
}