传送门

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4527    Accepted Submission(s): 1869 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   Author GTmac   Source 2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT   Recommend zhouzeyong

直观想法： 所有小于n且与n为非互质数和=所有小于n数的和-所有小于n且与n互质的数的和

  原理：  求所有小于N且与N为互质数的和：         1.欧拉函数可求与N互质的数的个数 2.if gcd(n,i)=1 then gcd(n,n-i)=1 (1<=i<=n) 若已知m与n互质，则n-m也与n互质  那么，对于任何一个i与n互质，必然n-i也和n互质，所以PHI（N）必然是偶数（除了2） 所以从1到N与N互质的数的和为PHI(N)*N/2   #include <cstdio> #include <cstdlib> #include <ctime> #include <algorithm> typedef long long ll; const int s = 10, MAX_F = 70; const int mod=1e9+7; int main() { ll n; while (scanf("%lld", &n), n) { ll x=n; if(n==0) break; ll ans=n; if(n%2==0)//只搜n/2次，不然可能超一组 （也可能不超） { while(n%2==0) n/=2; ans=ans/2; } for(long long i=3; i*i<=n; i+=2) //每找到一个就更改上界，这个优化很多，所以直接copy刘汝佳的代码会超时 { if(n%i==0) { while(n%i==0) n/=i; ans=ans/i*(i-1); } } if(n>1) ans=ans/n*(n-1); ll sum=(x-1)*x/2; sum%=mod; sum=sum-(ans*x/2)%mod; sum=(sum+mod)%mod; printf("%lld\n",sum); } return 0; }