POJ 3604 Professor Ben

Professor Ben

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8995		Accepted: 2298

Description

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_nare 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

Source

POJ Founder Monthly Contest – 2008.06.29, Lin Ben

題意：給一個數，求這個數的因數的因數個數的立方和，比如6的因子有1 2 3 6，1的因子數爲1，2的爲2，3的爲2，6的爲4，所以答案就是1^3+2^3+3^3+4^3=81

我是通過找規律找到答案的，如f(6)=f(2)*f(3)=81,而f(2)=f(3)=9,即當m，互質時f(m*n)=f(m)*f(n),當m可化爲n^p時，f(m)=f(n^p)=(1^3+2^3+3^3+……+p^3),如f(4)=f(2^2)=1^3+2^3+3^3,f(18)=f(3^2*2)=f(3^2)*f(2)=36*9,當n爲質數時,f(n)=9,結論就很明顯了，就是一個唯一分解

詳細推導過程可參考：http://blog.csdn.net/famousdt/article/details/7285382

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
const int Max=3e3;
int m,n,p[Max/5];
ll a[200],ans;
bool vis[Max+1];

void init()//打立方表，2^23>5e6,
{
    a[1]=1;
    for(int i=2;i<=23;i++)
        a[i]=a[i-1]+(ll)i*i*i;
}

void prime()//打素數表到3e3
{
    mem(vis,0);
    p[0]=0;
    vis[1]=1;
    for(int i=2;i<=Max;i++)
    {
        if(!vis[i])p[++p[0]]=i;
        for(int j=1;j<=p[0]&&(ll)i*p[j]<=Max;j++)
        {
            vis[i*p[j]]=1;
            if(!(i%p[j]))break;
        }
    }
}

void solve()//唯一分解
{
    for(int i=1;i<=p[0]&&(ll)p[i]*p[i]<=n;i++)
    {
        int num=0;
        while(!(n%p[i]))
            num++,n/=p[i];
        if(num>1)
            ans*=a[num+1];
        else if(num)
            ans*=9;
    }
    if(n>1)ans*=9;
}

int main()
{
    init();
    prime();
    sf("%d",&m);
    while(m--)
    {
        ans=1;
        sf("%d",&n);
        solve();
        pf("%lld\n",ans);
    }
    return 0;
}