hdoj-3501-Calculation 2-欧拉函数

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2451    Accepted Submission(s): 1023

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

340

 

Sample Output

02题目大意为：给一个整数n,求所有大于1小于n的数中，与n不是互质的数  的和例如：n=6,  
大于1小于6的数中，与6不是互质的数有2,3,4其和为9

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<queue>#include<stack>using namespace std;__int64 euler(__int64 x){__int64 res=x;for(__int64 i=2;i<=floor(sqrt(x*1.0));i++){if(x%i==0){res-=res/i;while(x%i==0){x/=i;} }} if(x>1) res-=res/x;return res;}int main(){__int64 n;while(~scanf("%I64d",&n),n)//(1 ≤ N ≤ 1000000000).  输入为何要用64??? {__int64 ans=(n*(n-1)/2-n*euler(n)/2)%1000000007;printf("%I64d\n",ans);} return 0;}