Longge's problem
 

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
思路:比如一个数n;
   n跟一个数的gcd为k;
   即求phi(n/k);
 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
int res = , ch ;
while( !( ( ch = getchar() ) >= '' && ch <= '' ) )
{
if( ch == EOF ) return << ;
}
res = ch - '' ;
while( ( ch = getchar() ) >= '' && ch <= '' )
res = res * + ( ch - '' ) ;
return res ;
}
#define MAXN 100001
ll prime[MAXN];//保存素数
ll vis[MAXN],ji;//初始化
ll Prime(ll n)
{
ll cnt=;
//memset(vis,0,sizeof(vis));
for(ll i=;i<=n;i++)
{
if(!vis[i])
prime[cnt++]=i;
for(ll j=;j<cnt&&i*prime[j]<n;j++)
{
vis[i*prime[j]]=;
if(i%prime[j]==)//关键
break;
}
}
return cnt;
}
ll phi(ll n)
{
ll i,rea=n;
for(i=;i<ji;i++)
{
if(prime[i]*prime[i]>n)break;
if(n%prime[i]==)
{
rea=rea-rea/prime[i];
while(n%prime[i]==) n/=prime[i];
}
}
if(n>)
rea=rea-rea/n;
return rea;
}
int main()
{
ll x,y,z,i,t;
ji=Prime();
while(~scanf("%I64d",&x))
{
ll ans=;
for(i=;i*i<=x;i++)
{
if(x%i==)
{
ll gg=i;
ll hh=x/i;
ans+=gg*phi(hh);
if(hh!=gg)
ans+=hh*phi(gg);
}
}
printf("%I64d\n",ans);
}
return ;
}

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