POJ2480 Longge's problem

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7994		Accepted: 2650

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

 

和BZOJ2705一样的题……

本来觉着把之前的代码复制过来稍微改改就能过了，然而做不到！

各种TLE！

把能想到的优化都堆上去了，好不容易才AC。

POJ数据是有多猛……

 

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
inline long long phi(long long x){
    long long a=x;
    long long m=sqrt(x+0.5);
    for(long long i=2;i<=m;i++){
        if(x%i==0){//找到因数 
            a=a/i*(i-1);//基本计算公式 a*=((i-1)/i) 
            while(x%i==0)x/=i;//除去所有相同因数 
        }
    }
    if(x>1)a=a/x*(x-1);//处理最后一个大因数 
    return a;
}
int main(){
    long long n;
    while(~scanf("%lld",&n)){
        long long ans=0;
        long long m;
        m=floor(sqrt(1.0*n));
        long long i;
        for(i=1;i<=m;++i){
            if(n%i==0){
                ans+=(long long)phi(n/i)*i;
                if(i*i<n)ans+=(long long)(n/i)*phi(i);
            }
        }
        printf("%lld
",ans);
    }
    return 0;
}