HDU 3501 Calculation 2(欧拉函数的引申）

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1264    Accepted Submission(s): 530

Problem Description

Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.

 

 

Input

For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.

 

 

Output

For each test case, you should print the sum module 1000000007 in a line.

 

 

Sample Input

3 4 0

 

 

Sample Output

0 2

 

 

Author

GTmac

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

 

 

Recommend

zhouzeyong

 

 

 

其实就是一个欧拉函数的推广。

小于等于n,与n互质的数的个数是phi(n) 叫欧拉函数

小于等于n,与n互质的数的和是  phi(n)*n/2;

所以总和减掉就是答案了。

/*

HDU 3501

求小于N与N不互质的数的和

欧拉公式的引伸：小于或等于n的数中，与n互质的数的总和为：φ(x) * x / 2。(n>1)





*/

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;

const int MOD=1000000007;



//求欧拉函数

long long eular(long long n)

{

    long long ret=n;

    long long i;

    for(i=2;i*i<=n;i++)

    {

        if(n%i==0)

        {

            ret-=ret/i;

            while(n%i==0)n/=i;

            if(n==1)break;

        }

    }

    if(n>1)ret-=ret/n;

    return ret;

}

int main()

{

    long long n;

    while(scanf("%I64d",&n),n)

    {

        long long ans=(n*(n-1)/2%MOD-eular(n)*n/2%MOD+MOD)%MOD;

        printf("%I64d\n",ans);

    }

    return 0;

}