hdu 3501 Calculation 2

Calculation 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2570 Accepted Submission(s): 1073



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

题意：告诉一个数n,求 1到n之中与n不互质的数的和

思路：先求互质的和，在用前n-1个数的和来减。

此处用到欧拉函数。（对于一个数n，在小于n的数中与n互质的数的个数）

一个关于gcd的定理。gcd(n,i)=1，那么gcd(n,n-i)=1，互质的所有数的和，sum=(eular(n)*n/2) (n-i与i和为n，个数为eular/2个）

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 1000000001
#define mod 1000000007
using namespace std;

typedef long long ll;

ll eular(ll n)
{
    ll num=1;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            num*=(i-1);
            n/=i;
            while(n%i==0)
            {
                n/=i;
                num*=i;
            }
        }
    }
    if(n>1) num*=(n-1);

    return num;
}

int main()
{
    ll n;
    while(scanf("%d",&n),n)
    {
        ll ans;
        ans=n*(n+1)/2-n;

        ans-=(eular(n)*n/2);//由欧拉函数求出与n互质的数，减去互质数的和
        printf("%I64d\n",(ans%mod+mod)%mod);
    }

    return 0;
}