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=2;i*i<=n;i++)
    if(n%i==0)
    {
        while(n%i==0)
        n=n/i;
        temp=temp/i*(i-1);
    }
    if(n!=1)
    temp=temp/n*(n-1);
    return temp;
}

int main()
{
    __int64 n,i,sum,k;
    while(scanf("%I64d",&n)>0)
    {
       sum=0;
       for(i=1;i*i<=n;i++)
       {
           if(n%i==0)
           sum=sum+Euler(n/i)*i;
           k=n/i;
           if(n%k==0 && k!=i)
           sum=sum+Euler(n/k)*k;
       }
       printf("%I64d
",sum);
    }
    return 0;
}