POJ2081：Recaman's Sequence

Description

Input

Output

Sample Input

7
10000
-1

Sample Output

20
18658

题目意思不难理解就是第m个位置的数是根据第m-1位置的数推出来的如果a[m-1]-m>0，并且a[m-1]-m在前面的序列中没有出现过那么a[m] = a[m-1]-m否则a[m] = a[m-1]+m

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[500005]={0};
bool hash[10000000]={false};

int main()
{
    int i;
    for(i = 1;i<= 500000;i++)
    {
        if(a[i-1]-i>0 && !hash[a[i-1]-i])
        a[i] = a[i-1]-i;
        else
        a[i] = a[i-1]+i;
        hash[a[i]] = true;//hash[a[i]] = true,标志为已经出现过
    }
    while(~scanf("%d",&i),i>=0)
    printf("%d\n",a[i]);

    return 0;
}

POJ2081：Recaman's Sequence,布布扣,bubuko.com

POJ2081：Recaman's Sequence

原文：http://blog.csdn.net/libin56842/article/details/20237999