POJ 2081——Recaman's Sequence

Recaman's Sequence

Time Limit: 3000MS Memory Limit: 60000KTotal Submissions: 22885 Accepted: 9881

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

题意：已知a[0]=0;如果a[m]-m>0并且a[m]-m在数列中没有出现过，则a[m]=a[m-1]-m,反之a[m]=a[m-1]+m.

解：直接打表操作，使用map来进行标记该数是否在数列中出现过。

#include<stdio.h>#include<map>#include<iostream>using namespace std;map<int ,int> m;int a[500005];void f(){a[0]=0;m[a[0]]++;for(int i=1;i<500005;i++){int x=a[i-1]-i;if(m[x]==0 && x>0){a[i]=a[i-1]-i;m[x]++;}else {a[i]=a[i-1]+i;m[a[i]]++;}}}int main(){int n;f();while(~scanf("%d",&n) && n>=0){printf("%d\n",a[n]);}}