Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3791    Accepted Submission(s): 1587 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

#include<cstdio> typedef long long LL; #define M 1000000007 int main() { LL n; while(~scanf("%lld",&n),n) { LL i,ans=n,temp=n,sum1,sum2; for(i=2;i*i<=n;++i) { if(n%i==0) ans=ans*(i-1)/i; while(n%i==0) n/=i; } if(n!=1) ans=ans*(n-1)/n; sum1=ans*temp/2;//计算所有和n互质的数的和 sum2=temp*(temp-1)/2;//求出1-n-1的和 //printf("%lld\n",sum1); //printf("%lld\n",sum2); printf("%lld\n",(sum2-sum1)%M); } return 0; }