POJ2081 Recaman's Sequence

PS:  此题目可以用DP解决，如果打表时间会很快。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <set>
#include <algorithm>
using namespace std;
const int maxn = 500100;
//Accepted	19140K	2454MS	C++	648B	2014-04-10 11:11:04
int d[maxn];
set<int> s;

void dp() {
    s.clear();
    s.insert(0);
    d[0] = 0;
    for(int i = 1; i <= 500001; i++) {
        d[i] = d[i-1] - i;
        if(d[i]>0 && s.count(d[i])==0) {
            s.insert(d[i]);
            continue;
        }
        else {
            d[i] = d[i-1] + i;
            s.insert(d[i]);
        }
    }
}


int main()
{
    dp();
    int n;
    while(scanf("%d", &n)==1 && n!=-1) {
        printf("%d\n", d[n]);
    }
    return 0;
}

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POJ2081 Recaman's Sequence

原文：http://blog.csdn.net/achiberx/article/details/23341087