HDU 3501 Calculation 2（欧拉函数）

Calculation 2

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

简单翻译：

输入一个n，求小于n，并且不与n互质的数的和模1000000007的值。

 

解题思路：

欧拉函数，先求出小于n且与n互质的数的和，用总数减去互质的数的和，就得到了不互质的数的和。

 

代码：

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<iostream>

 4 using namespace std;

 5 const int mod=1000000007;

 6 int main()

 7 {

 8     int n;

 9     while(scanf("%d",&n)!=EOF&&n)

10     {

11         long long temp=n,eular=1;

12         long long ans=(temp*(temp+1)/2)%mod;

13         for(int i=2;i*i<=temp;i++)

14             if(temp%i==0)

15             {

16                 temp/=i;

17                 eular*=i-1;

18                 while(temp%i==0)

19                 {

20                     temp/=i;

21                     eular*=i;

22                 }

23             }

24         if(temp>1) eular*=temp-1;

25         eular%=mod;

26         temp=n;

27         long long w=(eular*temp/2)%mod;

28         ans-=w;

29         ans-=n;

30         while(ans<0) ans+=mod;

31         printf("%lld\n",ans);

32     }

33     return 0;

34 }

Calculation 2