POJ 3604 Professor Ben

Professor Ben

Time Limit: 5000MS Memory Limit: 65536KTotal 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

14

Sample Output

36

Source

POJ Founder Monthly Contest – 2008.06.29, Lin Ben

题意：给一个数，求这个数的因数的因数个数的立方和，比如6的因子有1 2 3 6，1的因子数为1，2的为2，3的为2，6的为4，所以答案就是1^3+2^3+3^3+4^3=81

我是通过找规律找到答案的，如f(6)=f(2)*f(3)=81,而f(2)=f(3)=9,即当m，互质时f(m*n)=f(m)*f(n),当m可化为n^p时，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,当n为质数时,f(n)=9,结论就很明显了，就是一个唯一分解

详细推导过程可参考：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()//打素数表到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;}