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 ",ans); 32 } 33 return 0; 34 }