The problem of SDOI2012 Longge

Title Description Description

Longge is very good at maths and he is very happy to challenge difficult math problems. Here's the problem:
To set an integer N, you need to ask for ΣGCD (i, N) (1<=i<=n).

Enter a description Input Description

The first line contains an integer N, as shown in title

Output description Output Description

The first line contains an integer that is the answer to the request.

Sample input Sample Input

6

Sample output Sample Output

15

Data range and Tips Data Size & Hint

For 60% of data, 0<n<=2^16.
For 100% of data, 0<n<=2^32.

We set the number of GCD (i,n) =k I to A (K), so the answer is k1* (A (K1)) + K2 * (A (K2)) +...+ kp* (A (KP)), where K1 K2 ... KP is an approximate n of each. The number of 1-n that can be divisible by K is a total of n/k, and in this n/k number, some of the numbers gcd (i,n) =k, while some gcd (i,n) = CK (c>1,c∈r). Then we need to find out those numbers GCD (i,n) =k. First, the number K must meet GCD (k,n) =k, if you want k multiplied by a number can still satisfy the GCD (k,n) =k, then he multiply this number must be less than coprime and n/k, less than very good understanding. What I need to explain is why need coprime, take out a n/k factor (not including 1), get the number pk,n must divide the PK, so gcd (pk,n) is equal to k*p. We can use Euler functions to find the numbers less than and coprime. So, we just need to find out all the factors of N.

1#include <iostream>2#include <vector>3#include <cmath>4#include <cstring>5 using namespacestd;6 7vector<Long Long>Ys;8 9 Long LongN, ans;Ten  One  A Long LongPhiLong LongN) {//Euler functions -     if(n = =1)return 1; -     Long LongK =N; the      for(inti =2; I*i <= N; i++) -         if(n% i = =0)  -         { -K/=i; +              while(n% i = =0) n/=i; -K *= I-1; +         } A     if(N >1) { atK/=N; -K *= (N-1); -     } -     returnK; - } -  in intMain () { -CIN >>N; to     intnn = floor (sqrt (n));//Upper Bounds +      for(inti =1; I <= nn; i++)//find the Factor -         if(n% i = =0) the             if(I*i = =N) ys.push_back (i); *             Else { $ Ys.push_back (i);Panax NotoginsengYs.push_back (n/i); -             } the      for(inti =0; I < ys.size (); i++) { +Ans + = ys[i] * PHI (N/ys[i]); A     } thecout << ans <<"\ n"; +     return 0; -}

The problem of SDOI2012 Longge