HDU 3501 Calculation 2------欧拉函数变形

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1750    Accepted Submission(s): 727

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

 /*
 欧拉函数一个小小的性质：

 求小于n并且与n互质的数的和为：n*phi[n]/2；

 这个好像以前有点印象的。

 */

 #include<stdio.h>

 __int64 Euler(__int64 n)
 {
     __int64 temp=n,i;
     for(i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             while(n%i==)
             n=n/i;
             temp=temp/i*(i-);
         }
     }
     if(n!=) temp=temp/n*(n-);
     return temp;
 }

 int main()
 {
     __int64 n,m;
     __int64 k;
     while(scanf("%I64d",&n)>)
     {
         if(n==)break;
         m=Euler(n);
         k=m*n/;
         k=n*(n+)/-n-k;
         //刚开始，只有k为__in64，错了。
         //这里相乘之后，可能会超过int范围。倒置溢出
         k=k%;
         printf("%I64d\n",k);
     }
     return ;
 }