poj 2480 Longge's problem 欧拉函数+素数打表

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 = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    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=0;
    //memset(vis,0,sizeof(vis));
    for(ll i=2;i<=n;i++)
    {
        if(!vis[i])
        prime[cnt++]=i;
        for(ll j=0;j<cnt&&i*prime[j]<n;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)//关键
            break;
        }
    }
    return cnt;
}
ll phi(ll n)
{
    ll i,rea=n;
    for(i=0;i<ji;i++)
    {
        if(prime[i]*prime[i]>n)break;
        if(n%prime[i]==0)
        {
            rea=rea-rea/prime[i];
            while(n%prime[i]==0)  n/=prime[i];
        }
    }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}
int main()
{
    ll x,y,z,i,t;
    ji=Prime(52010);
    while(~scanf("%I64d",&x))
    {
        ll ans=0;
        for(i=1;i*i<=x;i++)
        {
            if(x%i==0)
            {
                ll gg=i;
                ll hh=x/i;
                ans+=gg*phi(hh);
                if(hh!=gg)
                ans+=hh*phi(gg);
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}