Calculation of (number theory) numbers

Title Description Description

We need to find out the number of characters with the following properties (the natural number of inputs N):

First enter a natural number n (n<=1000), and then treat this natural number as follows:

1. Do not make any treatment;

2. Add a natural number to the left of it, but the natural number cannot exceed half of the original number;

3. After adding a number, continue to follow this rule until no more natural numbers can be added.

Enter a description Input Description

A number n

Output description Output Description

Number of numbers that satisfy the condition

Sample input Sample Input

6

Sample output Sample Output

6

Data range and Tips Data Size & Hint

The 6 numbers were:

6

16

26

126

36

136

Analysis:

1. The difficulty seems to be small, but it takes a lot of time to do it by hand, because the amount of data that needs to be repeated is too large. Of course, we can also take the side of the method of recursive side storage, but the amount of computation is still not small, and then further thinking, in fact, can be used as the following recursive method to do;
2. For example, F (6) is required, after analysis, we know that: f (6) =f (1) +f (2) +f (3) +1, that is, the number of answers to F (6) is the sum of the number of answers to all the natural numbers that can be taken before it (6 can be taken by three natural numbers), The last plus 1 means that the number 6 itself is also an answer;
3. So, we can know F (n) =f (1) +f (2) +......f (Trunc (N/2)) +1;
4. Therefore, F (n) is required, we simply use the above formula to compile a recursive process, the F (2) to f (n) All can be obtained, for F (1000) and no more than 1 seconds to obtain the results.
The second type of algorithm:
1. For f (7) =f (6) is obvious, namely F (2n+1) =f (2n). So what is the relationship between F (8) and F (7)?
2. The analysis shows that the difference between f (8) and F (7) is that F (8), in addition to all cases containing F (7), adds an additional F (4), namely: F (8) =f (7) +f (4). Therefore available: f (2n) =f (2n-1) +f (n). Simply by compiling a recursive small process or by using a recursive method.

1#include <iostream>2 using namespacestd;3 intMain ()4 {5     inti,n,ans,sum[1001];6sum[0] =0, Sum [1] =1;7Cin>>N;8      for(i =2; I <= N; i++)9     {TenAns = sum[i/2] +1; Onesum [i] = sum[i-1] +ans; A     } -Cout<<sum[n]-sum[n-1]<<Endl; -     return 0; the}

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Calculation of (number theory) numbers