Untitled---------------Constructors

Title Description

BobGot aAliceThe gift was very happy and studied more numbers.BobThink a numberX is graceful when and only if there is a positive integerAand a> 1  positive integers   b, making  x = A^b < Span class= "Mord mathit mtight".

Bob < Span class= "Mord mathit" >  like to count these beautiful numbers when it's okay, but  bo b  has limited computational power if the numbers are too big to come out. Now he wants you to help him figure out  1< Span class= "Strut bottom" >  to  n< The number of spans class= "Mord mathit" >  is graceful.

Input/output format

Input format:

One line is a positive integer N, as described in title.

Output format:

A row of an integer anS, representing the number of graceful numbers in 1 1 to N.

Input and Output Sample input example # #:

10

Sample # # of output:

4

Description

There are 1, 4, 8, 9 within 1 0 of the graceful.

The data for 30% satisfies   1 ≤  n ≤  10 < Span class= "Sizing reset-size6 size3 mtight" >5.

The data for 70% satisfies   1 ≤  n ≤  10 < Span class= "Sizing reset-size6 size3 mtight" >10.

The data for 100% satisfies 1 ≤ N ≤ 1018.

100-minute Procedure:

< Span class= "Mord" > < Span class= "vlist-t" > 1. 

< Span class= "Mord" > < Span class= "vlist-t" > 2. For a contributing X, we use all x = A λb to represent it in the largest form of B.

< Span class= "Mord" > < Span class= "vlist-t" > 3. The definition function g (b) indicates how many x can be represented as a λb. The function f (b) indicates how many x can be expressed as aλb, and B satisfies the above properties.

< Span class= "Strut bottom" > < Span class= "Mord mtight" > 4. Obviously the and of the F function is the answer.

< Span class= "Strut bottom" > < Span class= "Mord mtight" > 5. G function: Pow (n, 1/b).
/span>

< Span class= "Strut bottom" > < Span class= "Mord mtight" > 6. The method of the f  function: g (b)-σi | b f (i).

1#include <bits/stdc++.h>2 #definell Long Long3 using namespacestd;4ll n,f[ -],ans;5 intMain ()6 {7scanf"%lld",&n);8      for(intI= -; i>=2;--i)9     {Tenf[i]= (LL) POW (n,1.0/i)-1; One          for(intj=i+i;j<= -; j+=i) Af[i]-=F[j]; -ans+=F[i]; -     } theprintf"%lld", ans+1); -     return 0; -}

Code

Untitled---------------Constructors