Professor Ben 唯一分解定理：因子的因子個數

題意：一個數N,它的因子爲a1，a2....ai...an,求每一個因子的因子個數的三次方；

例如4的因子1,2,4。1的因子個數爲1,2的因子個數爲2，4的因子個數爲3

結果bns=1^3+2^3+3^3=36;

思路：唯一分解定理（自己去網上了解）求每個素數因子的指數

例如36=2^2*3^2;2的指數是2，意味着有三種組合，2^0,2^1,2^2,

每一種組合能提供（1+0）,(1+1),(1+3)個因子，同理可得3^2;

所以答案爲(（1+0）^3+（1+1）^3（1+2）^3)*(（1+0）^3+（1+1）^3（1+2）^3);

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
const int N=1e4+5;
int tot=0,ans[N];
bool vis[N];
ll power[25];

void primer()
{
    memset(vis,true,sizeof(vis));
    for(int i=2;i<N;i++)
    {
        if(vis[i]) ans[++tot]=i;
        for(int j=1;(j<=tot)&&(ans[j]*i<N);++j)
        {
            vis[ans[j]*i]=false;
            if(i%ans[j]==0) break;
        }
    }
}

void powe()
{
    for(int i=1;i<23;i++)
        for(int j=0;j<=i;j++)
        power[i]+=(1+j)*(1+j)*(1+j);
}

ll pre(int n)
{
    ll bns=1;
    for(int i=1;ans[i]*ans[i]<=n;i++)
    {
        int cnt=0;
        while(n%ans[i]==0){
            n/=ans[i];
            cnt++;
        }
        if(cnt>0) bns*=power[cnt];
    }
    if(n>1)  bns*=9;
    return bns;
}

int main()
{
    powe();
    primer();
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%lld\n",pre(n));
    }
    return 0;
}

題目：

Professor Ben is an old stubborn man teaching mathematics in a university. He likes to puzzle his students with perplexing (sometimes boring) problems. Today his task is: for a given integer N, a₁,a₂, ... ,a_n are the factors of N, let b_i be the number of factors of a_i, your job is to find the sum of cubes of all b_i. Looking at the confused faces of his students, Prof. Ben explains it with a satisfied smile:

Let's assume N = 4. Then it has three factors 1, 2, and 4. Their numbers of factors are 1, 2 and 3 respectively. So the sum is 1 plus 8 plus 27 which equals 36. So 36 is the answer for N = 4.

Given an integer N, your task is to find the answer.

Input

The first line contains the number the test cases, Q(1 ≤ Q ≤ 500000). Each test case contains an integer N(1 ≤ N ≤ 5000000)

Output

For each test case output the answer in a separate line.

Sample Input

1
4

Sample Output

36