HDU3501 Calculation 2 (Euler function)

Calculation 2Time limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U SubmitStatus

Description

Given a positive integer n, your task is to calculate the sum of the positive integers less than N which was not coprime t o N. A is said to being coprime to B if A, B share no common positive divisors except 1.

Input

For each test case, the 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 of module 1000000007 in a line.

Sample Input

340

Sample Output

02 Simple translation:Enter an n, which is less than N and does not match the number of n coprime with the value of modulo 1000000007. Problem Solving Ideas:Euler function, the sum of the number of less than N and n coprime is calculated, and the sum of the number minus coprime is obtained, and the sum of the number is not coprime. Code:

1#include <cstdio>2#include <cstring>3#include <iostream>4 using namespacestd;5 Const intMod=1000000007;6 intMain ()7 {8     intN;9      while(SCANF ("%d", &n)!=eof&&N)Ten     { One         Long LongTemp=n,eular=1; A         Long LongAns= (temp* (temp+1)/2)%MoD; -          for(intI=2; i*i<=temp;i++) -             if(temp%i==0) the             { -Temp/=i; -eular*=i-1; -                  while(temp%i==0) +                 { -Temp/=i; +eular*=i; A                 } at             } -         if(temp>1) eular*=temp-1; -eular%=MoD; -temp=N; -         Long LongW= (eular*temp/2)%MoD; -ans-=W; inans-=N; -          while(ans<0) ans+=MoD; toprintf"%lld\n", ans); +     } -     return 0; the}

Calculation 2

HDU3501 calculation 2 (Euler function)