UVa 10049 self-describing Sequence: self-describing sequence & binary recursion

10049-self-describing Sequence

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=34&page=show_problem &problem=990

Solomon Golomb ' s selfdescribing sequence is the only nondecreasing sequence of positive It contains exactly f (k) occurrences of K for each k. A few moments thought reveals that sequence must begin as follows:

In this problem your are expected to write a program that calculates the value of f (n) Given the value of N.

Input

The input may contain multiple test cases. Each test case occupies a separate line and contains an integer n (). The input terminates with a test case containing a value of 0 for N and the must is processed.

Output

For each test case in the input output the value of f (n) on a separate line.

Sample Input

9999
123456
1000000000
0

Sample Output

356
1684
438744

1. How do I construct a recursive type with a faster growth rate?

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45359.htm

Note that f (n) grows slowly, and you can get f (n) by using its inverse function g (n) =max{m | f (M) =n} (such as G (4) =8), and then the inverse function f (n) =min{k | g[k]>=n} is obtained.

Then the definition of f (n) has g (n) =g (n-1) +f (n) =g (n-1) +min{k | g[k]>=n}

(e.g. G (4) =g (3) +f (4) =5+3=8,g (5) =g (4) +f (5) =8+3=11)

2. How to calculate Min{k | g[k]>=n}?

Use a binary lookup.

Complete code:

/*0.075s*/
    
#include <cstdio>  
#include <algorithm>  
const int M = 700000;  
using namespace std;  
    
Long long g[m];  
    
int main ()  
{  
    g[1] = 1;  
    G[2] = 3;  
    for (int i = 3; i < M ++i)  
        g[i] = g[i-1] + (Lower_bound (g + 1, g + I, i)-g);  
    int n;  
    while (scanf ("%d", &n), N)  
        printf ("%d\n", Lower_bound (g + 1, G + M, N)-G);  
    return 0;  
}