poj2081 Recaman's Sequence

来源:互联网 发布:水彩 水粉 油画 知乎 编辑:程序博客网 时间:2024/05/16 15:22

C - Recaman's Sequence
Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:60000KB     64bit IO Format:%I64d & %I64u
SubmitStatus Practice POJ 2081
Appoint description:

Description

The Recaman's sequence is defined by a0 = 0 ; for m > 0, am = a m−1 − m if the rsulting a m is positive and not already in the sequence, otherwise am = a m−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 a k.

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

710000-1

Sample Output

2018658
打表输出即可。

#include <stdio.h>#include <string.h>bool vis[6555550];int result[500000+5];int main(){memset(vis,false,sizeof(vis));result[0]=0;vis[0]=true;for(int i=1;i<=500000;i++){int x=result[i-1]-i;if(x>0&&!vis[x]){result[i]=x;vis[x]=true;}else{result[i]=result[i-1]+i;vis[result[i]]=true;}}int n;while(~scanf("%d",&n)&&n!=-1){printf("%d\n",result[n]);}return 0;}



0 0
原创粉丝点击