HDOJ Calculation 2 3501(eular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): 2511 Accepted Submission(s): 1048
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
Author
GTmac
Source
2010 ACM-ICPC Multi-University Training Contest(7)——Host by HIT
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欧拉函数性质:N>1,不大于N且和N互素的所有正整数的和是 1/2*M*eular(N)。
#include<stdio.h>#include<algorithm>using namespace std;typedef __int64 LL;LL phi(LL n){LL i,res=n;for(i=2;i*i<=n;i++) if(n%i==0){ res=res/i*(i-1); while(n%i==0)n/=i;}if(n>1) res=res/n*(n-1);return res;}int main(){LL N;while(scanf("%I64d",&N),N){LL x=(N*(N-1)/2-N*phi(N)/2)%1000000007;printf("%I64d\n",x);}return 0;}
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