HDU 3501 Calculation 2

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1296    Accepted Submission(s): 544

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

N以内与N互质的数和 的扩展

N以内与N互质的数和 =》phi[N]*N/2 
而本题是求与N不互质的和，就很简单
#include <iostream>

#include <map>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

#define N 32000

bool h[N];

int p[4000];



void prime()//帅选求素数

{

    int cnt=1;

    int i,j;

    p[0]=2;

    for(i=2;i<N;i+=2)

      h[i]=1;

    for(i=3;i<N;i+=2)

     if(!h[i])

     {

         p[cnt++]=i;

        for(j=i*i;j<N;j+=i)

          h[j]=1;

     }

}

__int64 solve(__int64  n)

{

    int m=(int)sqrt(n*1.0);

    __int64 r=n,tp=n;

    int i;

    for(i=0;p[i]<=m;i++)

     if(n%p[i]==0)

       {

           tp=tp/p[i]*(p[i]-1);

           while(n%p[i]==0) n/=p[i];

           if(n==1) break;

       }



    if(n>1) tp=tp/n*(n-1);

    return (r*tp)>>1;

}

int main()

{

    prime();

    __int64 n,sum;

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

    {

        sum=((n-1)*n)>>1;

        sum-=solve(n);

        printf("%I64d\n",sum%1000000007);

    }

    return 0;

}