HDOJ 题目3501 Calculation 2（欧拉函数）

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Calculation 2

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2288 Accepted Submission(s): 971

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

Sample Output

Author

GTmac

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

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欧拉函数是算小于n的且与n互素的个数，当n与a互斥时，n必定与n-a互斥，所以总共会有eular（n)/2对，乘以n就是互素的数的和，前n-1项减去就行

ac代码

#include<stdio.h>#include<string.h>#define mod 1000000007__int64 eular(__int64 n){__int64 ans=n,i;for(i=2;i*i<=n;i++){if(n%i==0){ans-=ans/i;while(n%i==0)n/=i;}}if(n>1){ans-=ans/n;}return ans;}int main(){__int64 n;while(scanf("%I64d",&n)!=EOF,n){printf("%I64d\n",(n*(n-1)/2-eular(n)*n/2)%mod);}}

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