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

 1 /*

 2 欧拉函数一个小小的性质：

 3 

 4 求小于n并且与n互质的数的和为：n*phi[n]/2；

 5 

 6 这个好像以前有点印象的。

 7 

 8 */

 9 

10 #include<stdio.h>

11 

12 

13 __int64 Euler(__int64 n)

14 {

15     __int64 temp=n,i;

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

17     {

18         if(n%i==0)

19         {

20             while(n%i==0)

21             n=n/i;

22             temp=temp/i*(i-1);

23         }

24     }

25     if(n!=1) temp=temp/n*(n-1);

26     return temp;

27 }

28 

29 int main()

30 {

31     __int64 n,m;

32     __int64 k;

33     while(scanf("%I64d",&n)>0)

34     {

35         if(n==0)break;

36         m=Euler(n);

37         k=m*n/2;

38         k=n*(n+1)/2-n-k;

39         //刚开始，只有k为__in64，错了。

40         //这里相乘之后，可能会超过int范围。倒置溢出

41         k=k%1000000007;

42         printf("%I64d\n",k);

43     }

44     return 0;

45 }