The problem of Bzoj2705 Longge

Time Limit: 3000MS

Memory Limit: 131072KB

64bit IO Format: %lld &%llu

Description

Longge is very good at maths and he is very happy to challenge difficult math problems. Now the problem is: Given an integer N, you need to ask for ∑GCD (i, N) (1<=i <=n).

Input

an integer that is n.

Output

an integer, for the answer that is asked.

Sample Input

6

Sample Output

15

Hint

"Data Range"

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

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

Source

SDOI2012

Seeking ΣGCD (I,n) (1<=i<=n)

The brute force enumeration is certainly feasible, but tle cannot be avoided.

Considering the conversion problem, from 1 to N, there are many gcd of I (i,n) equal to the same n factor. We can enumerate each factor k of N, and accumulate "the number of gcd (i,n) with the number k as solution multiplied by the number s (k)" To get the answer.

If there is gcd (n,m) =k, then N and M with the exception of the Convention number k, you can get gcd (n/k,m/k) = 1. M/k and (n/k) coprime are known by the preceding formula. The number of (m/k) satisfies the condition, that is, S (k) is equal to Phi (n/k) ← Euler Function!

Solution 1: Direct set of templates. The algorithm is correct, but because the problem data is large, save the function after the re-processing will (the reason for the visual storage with the array can not open so large)

1 /*by Silvern*/2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cstdio>6#include <cmath>7 using namespacestd;8 Const intmaxn=100000;9 Long LongN;Ten intm[maxn],phi[maxn],p[maxn],pt; One intEuler () A { -phi[1]=1; -     intn=MAXN; the     intK; -      for(intI=2; i<n;i++) -     { -         if(!m[i])//I is the prime number +p[pt++]=m[i]=i,phi[i]=i-1; -          for(intj=0;j<pt&& (k=p[j]*i) <n;j++) +         { Am[k]=P[j]; at             if(M[i]==p[j])//in order to ensure that the future number is not re-screened, to break -             { -phi[k]=phi[i]*P[j]; - /*here Phi[k] and Phi[i] behind the ∏ (p[i]-1)/p[i] All the same (M[i]==p[j]) only one p[j], you can guarantee ∏ (p[i]-1)/p[i] in front of the same*/ -                  Break;  -             } in             Else -phi[k]=phi[i]* (p[j]-1);//properties of the integrable function, F (i*k) =f (i) *f (k) to         } +     } - } the intMain () { * Euler (); $scanf"%lld",&n);Panax Notoginseng     Long Longm=sqrt (n); -     inti,j; the     Long Longans=0; +      for(i=1; i<=m;i++){ A         if(n%i==0){ theans+=phi[n/i]*i; +ans+= (n/i) *Phi[i]; -         } $     } $printf"%lld\n", ans); -     return 0; -}

Solution 2: Take a little more time and count it all over again. And not tle, magical

The code was learned from HzW seniors, and it was streamlined.

1 /*by Silvern*/2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cstdio>6#include <cmath>7 using namespacestd;8 Const intmxn=100000;9 Long Longn,m;Ten Long LongPhiLong Longx) { One     Long LongA=x; A      for(Long LongI=2; i<=m;i++){ -         if(x%i==0){//find the Factor -a=a/i* (I-1);//Basic Calculation Formula a*= ((i-1)/i) the              while(x%i==0) X/=i;//Remove all the same factors -         } -     } -     if(x>1) a=a/x* (x1);//deal with the last big factor +     returnA; - } + intMain () { Ascanf"%lld",&n); atm=sqrt (n); -     inti,j; -     Long Longans=0; -      for(i=1; i<=m;i++){ -         if(n%i==0){ -Ans+=phi (n/i) *i; inans+= (n/i) *Phi (i); -         } to     } +printf"%lld\n", ans); -     return 0; the}

The problem of Bzoj2705 Longge