poj_2081

Recaman's Sequence

Time Limit: 3000MS Memory Limit: 60000KTotal Submissions: 19016 Accepted: 7963

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

DP水题

#include <iostream>#include <cstring>using namespace std;const int MAXN = 9999999;int a[MAXN];bool flag[MAXN];void DP(){for(int i=1;i<=500000;i++){if(a[i-1] - i > 0 && flag[a[i-1] - i] == false){a[i] = a[i-1] - i;flag[a[i-1] - i] = true;}else{a[i] = a[i-1] + i;flag[a[i-1] + i] = true;}}}int main(){memset(flag,false,sizeof(false));a[0] = 0;DP();int k;while(cin>>k){if(k == -1) break;cout<<a[k]<<endl;}}