[ACM_POJ_2081]动态规划入门练习（三）Recaman's Sequence

Recaman's Sequence

Time Limit: 3000MS Memory Limit: 60000K
Total Submissions: 17939 Accepted: 7450

Description
The Recaman's sequence is defined by a0 = 0 ; for m > 0, am = am−1 − m if the rsulting am is positive and not already in the sequence, otherwise am = am−1 + m.
The first few numbers in the Recaman's Sequence is 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9 ...
Given k, your task is to calculate ak.

Input
The input consists of several test cases. Each line of the input contains an integer k where 0 <= k <= 500000.
The last line contains an integer −1, which should not be processed.

Output
For each k given in the input, print one line containing ak to the output.

Sample Input
7
10000
-1

Sample Output
20
18658

Source

POJ2081

比较值得注意的是数组中已存在的数不再放入数组中，而数组很大（0 <= k <= 500000），因此定义另一个数组来储存该数是否已存在，以此避免查找，减少运行时间。

#include<stdio.h>
int f[500001];
bool b[10000000] = {1, 0};	//一个足够大的数组来储存是否已存在
int main(){
	f[0] = 0;
	for(int i = 1; i < 500001; ++i){
		int n = f[i - 1] - i;
		if(n < 0 || b[n]){
			n = f[i - 1] + i;
		}
		b[n] = 1;
		f[i] = n;
	}
	int m;
	while(scanf("%d", &m) && (m + 1)){
		printf("%d\n", f[m]);
	}
	return 0;
}

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