POJ 2081 Recaman's Sequence(哈希模拟)（水题）

最新推荐文章于 2019-06-16 13:08:20 发布

大白QQly成长日记

最新推荐文章于 2019-06-16 13:08:20 发布

阅读量594

点赞数

CC 4.0 BY-SA版权

分类专栏：基础经验文章标签：哈希

本文链接：https://blog.youkuaiyun.com/PleasantlY1/article/details/81391480

基础经验专栏收录该内容

27 篇文章

订阅专栏

本文介绍了Recaman数列的定义及生成算法，并提供了一个使用哈希表标记已出现数的C++实现方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：第0项为0，如果第m-1项-m>0且该数未出现过，第m项为第m-1项-m，否则为第m-1项+m；

思路：谁把这题放在dp专题的........太不靠谱了，hash表标记一下出现状态直接模拟。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[500005]={0};
int p[10000000]={0};
int main()
{
    int i;
    for(i=1;i<=500000;i++)
    {
        if(a[i-1]-i>0&&p[a[i-1]-i]==0)
        {
		   a[i]=a[i-1]-i;
	    }
        else{
        a[i]=a[i-1]+i;
        p[a[i]]=1;
        } 
    }
    while(~scanf("%d",&i),i>=0)
    printf("%d\n",a[i]);
    return 0;
}