POJ 2480 Longge's problem 欧拉函数—————∑gcd(i, N) 1<=i <=N

Longge's problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6383		Accepted: 2043

Description

Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N.

"Oh, I know, I know!" Longge shouts! But do you know? Please solve it.

Input

Input contain several test case. 
A number N per line. 

Output

For each N, output ,∑gcd(i, N) 1<=i <=N, a line

Sample Input

2
6

Sample Output

3
15

Source

POJ Contest,Author:Mathematica@ZSU

 /*

 题意：∑gcd(i, N) 1<=i <=N。由于N 2^31.
 刚开始想用欧拉求得1的个数，再求N的素因子，用容斥
 来求解。发现，就算是求得的素因子，也不最大公约数。

 HUD的一道题，提供了思路。

 枚举吧。
 if(N%i==0)
 {
 最大公约数为1的时候，有几个。欧拉值(N/1);
 最大公约数为2的时候，有几个。欧拉值(N/2);
 .....
 }

 */

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;

 __int64 Euler(__int64 n)
 {
     __int64 i,temp=n;
     for(i=;i*i<=n;i++)
     if(n%i==)
     {
         while(n%i==)
         n=n/i;
         temp=temp/i*(i-);
     }
     if(n!=)
     temp=temp/n*(n-);
     return temp;
 }

 int main()
 {
     __int64 n,i,sum,k;
     while(scanf("%I64d",&n)>)
     {
        sum=;
        for(i=;i*i<=n;i++)
        {
            if(n%i==)
            sum=sum+Euler(n/i)*i;
            k=n/i;
            if(n%k== && k!=i)
            sum=sum+Euler(n/k)*k;
        }
        printf("%I64d\n",sum);
     }
     return ;
 }