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Calculation 2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2570 Accepted Submission(s): 1073
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
题意：告诉一个数n,求 1到n之中与n不互质的数的和
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思路：先求互质的和，在用前n-1个数的和来减。
此处用到欧拉函数。（对于一个数n，在小于n的数中与n互质的数的个数）
一个关于gcd的定理。gcd(n,i)=1，那么gcd(n,n-i)=1，互质的所有数的和，sum=(eular(n)*n/2) (n-i与i和为n，个数为eular/2个）
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#include
#include
#include
#include
#include
#include
#define N 1000000001 #define mod 1000000007 using namespace std; typedef long long ll; ll eular(ll n) { ll num=1; for(int i=2;i*i<=n;i++) { if(n%i==0) { num*=(i-1); n/=i; while(n%i==0) { n/=i; num*=i; } } } if(n>1) num*=(n-1); return num; } int main() { ll n; while(scanf(%d,&n),n) { ll ans; ans=n*(n+1)/2-n; ans-=(eular(n)*n/2);//由欧拉函数求出与n互质的数，减去互质数的和 printf(%I64d ,(ans%mod+mod)%mod); } return 0; }
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