poj 2081 Recaman's Sequence

Recaman's Sequence

Time Limit: 3000MS Memory Limit: 60000KTotal Submissions: 22529 Accepted: 9678

Description

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

Sample Input

710000-1

Sample Output

2018658

Source

Shanghai 2004 Preliminary

#include<stdio.h>#include<string.h>int  res[500005];bool vis[3000005];//res[i]的结果会很大，要开大一点vis[] int main(){int n;memset(vis,true,sizeof(vis));res[0]=0;for(int i=1;i<=500000;i++){int x=res[i-1]-i;if(x>0&&vis[x]){ //判断当前值是否是正数并且保证不在res中 res[i]=res[i-1]-i;vis[x]=false; }else {res[i]=res[i-1]+i;vis[res[i]]=false;}}while(scanf("%d",&n)){if(n<0) break;printf("%d\n",res[n]);}return 0;}