HDU 3501 Calculation 2（欧拉函数）

Calculation 2

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

Submit Status

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：

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 using namespace std;
 const int mod=;
 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF&&n)
     {
         long long temp=n,eular=;
         long long ans=(temp*(temp+)/)%mod;
         for(int i=;i*i<=temp;i++)
             if(temp%i==)
             {
                 temp/=i;
                 eular*=i-;
                 while(temp%i==)
                 {
                     temp/=i;
                     eular*=i;
                 }
             }
         if(temp>) eular*=temp-;
         eular%=mod;
         temp=n;
         long long w=(eular*temp/)%mod;
         ans-=w;
         ans-=n;
         while(ans<) ans+=mod;
         printf("%lld\n",ans);
     }
     return ;
 }

Calculation 2