1236 - Pairs Forming LCM -- LightOj1236 (LCM)

http://lightoj.com/volume_showproblem.php?problem=1236

题目大意： 给你一个数n,让你求1到n之间的数（a,b && a<=b）两个数的最小公倍数等于n有多少对这样的ab.

分析都写在图片上了，费了我好大的事呢

ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>

using namespace std;
typedef long long LL;
#define N 10010001
#define ESP 1e-8
#define INF 0x3f3f3f3f
#define memset(a,b) memset(a,b,sizeof(a))

LL prime[], k;
bool vis[N];

void Prime()
{
    memset(vis, false);
    k = ;
    for(int i=; i<N; i++)
    {
        if(vis[i] == )
        {
            prime[k ++] = i;
            for(int j= i+i; j<N; j+=i)
            {
                vis[j] = ;
            }
        }
    }
}

LL solve(LL n)
{
    LL ans, sum;
    ans = ;
    sum = ;
    for(int i=; prime[i] * prime[i] <= n; i++)
    {
        if(n%prime[i] == )
        {
            ans=;
            while(n%prime[i] == )
            {
                ans ++;
                n /= prime[i];
            }
            sum *= (*ans+);
        }
    }
    if(n>)
        sum *= (* + );
    return sum;
}

int main()
{
    int T, t=;
    LL n;
    Prime();
    scanf("%d", &T);
    while(T --)
    {
        LL n;
        scanf("%lld", &n);

        LL sum = solve(n);

        printf("Case %d: %lld\n", t++, sum/+);
    }
    return ;
}