UVa 10049 Self-describing Sequence (自描述序列&二分递推)

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 integers with the property that it contains exactly f(k) occurrences of k for each k. A few moments thought reveals that the sequence must begin as follows:

In this problem you 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 0 for n and this case must not be processed.

Output

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

Sample Input

100
9999
123456
1000000000
0

Sample Output

21
356
1684
438744

1. 如何构造一个增长速率较快的递推式？

注意到f(n)增长缓慢，不妨利用其反函数g(n)=max{m | f(m)=n}(比如g(4)=8)，然后再求一次反函数f(n)=min{k | g[k]>=n}即可得到f(n)。

随后由f(n)的定义有g(n)=g(n-1)+f(n)=g(n-1)+min{k | g[k]>=n}

(比如g(4)=g(3)+f(4)=5+3=8，g(5)=g(4)+f(5)=8+3=11)

2. 如何计算min{k | g[k]>=n}？

用二分查找。

完整代码：

/*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;
}