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：

 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