POJ-2081-Recaman's Sequence

                            Time Limit: 3000MS      Memory Limit: 60000K                            Total Submissions: 22786        Accepted: 9821

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

题意说明：给你个数a[0]=0 当a[m-1]-m>0 并且这个式子求出来的数并未出现过，那么 am = am−1 − m,否则am = am−1 + m.

解题思路：这个不就不用多说了，为了节省时间，在输入前就应该遍历好这个数列，因为输出的数是不超过50W的，所有我们要遍历前50W个 中间要开个数组来记录出现过得数

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <map>#include <cmath>#include <queue>using namespace std;int k;int vis[5000005];//因为这50W中的数有些数很大 故要开这么大int dp[500005];//记录的数组int main(){    memset(vis,0,sizeof(vis));//初始化记录出现的数的数组    memset(dp,0,sizeof(dp));//初始化    dp[0]=0;//a[0]=0;    vis[dp[0]]=1;//标记0出现过    for(int i=1;i<=500000;i++)    {        if(dp[i-1]-i>0&&!vis[dp[i-1]-i]){//满足条件              dp[i]=dp[i-1]-i;              vis[dp[i]]=1;//记录出现过        }        else{            dp[i]=dp[i-1]+i;            vis[dp[i]]=1;//记录出现过        }    }    while(~scanf("%d",&k))    {        if(k==-1)          break;          printf("%d\n",dp[k]);    }    return 0;}

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