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Calculation 2(欧拉函数的运用)

时间:2016-05-04 01:10:26      阅读:213      评论:0      收藏:0      [点我收藏+]

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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
 
 
运用欧拉函数求与n互质的数的总和:phi[n]*n/2;
证明:在1<x<n中,与n互质的数有{x1,x2,x3....,xn},gcd(x,n)=gcd(n-x,n)=1,得:
{n-xn,....n-x2,n-x1},两个集合两两对应,相加除以2得phi[n]*n/2;
代码如下:
 
#include <stdio.h>
#include<math.h>
#include <string.h>


long long phi(long long n)
{
    long long res=n;
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
            {
                res=res-res/i;
                do
                    n/=i;
                while(n%i==0);
            }
    }
    if(n>1)
        res=res-res/n;
    return res;
}
int main()
{
    long long n,ans;
    while(scanf("%I64d",&n) != EOF)
    {
        if(n==0)
            break;
        ans=n*(n+1)/2-n;
        ans-=phi(n)*n/2;
        ans=ans%1000000007;
        printf("%I64d\n",ans);

    }


    return 0;
}
 
 

Calculation 2(欧拉函数的运用)

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原文地址:http://www.cnblogs.com/PLifer/p/5452573.html

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