POJ 2081 Recaman's Sequence

Recaman's Sequence

Time Limit: 3000ms

Memory Limit: 60000KB

This problem will be judged on PKU. Original ID: 2081
64-bit integer IO format: %lld      Java class name: Main

 

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

Source

Shanghai 2004 Preliminary

 

解题：离线搞。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <set>
#include <map>
#include <queue>
#include <ctime>
#define LL long long
#define INF 0x3f3f3f3f
#define pii pair<int,int>

using namespace std;
const int maxn = 500010;
int dp[maxn];
bool vis[5000000];
struct node {
    int k,ans,o;
};
node inp[1000000];
bool cmp1(const node &x,const node &y) {
    return x.k < y.k;
}
bool cmp2(const node &x,const node &y) {
    return x.o < y.o;
}
int main() {
    int tot = 0,tmp,cnt;
    for(int i = 1; i < maxn; ++i) {
        dp[i] = dp[i-1]-i;
        if(dp[i] <= 0 || vis[dp[i]])
            dp[i] = dp[i-1]+i;
        vis[dp[i]] = true;
    }
    while(~scanf("%d",&tmp)&&(~tmp)) {
        inp[tot].k = tmp;
        inp[tot].o = tot++;
    }
    sort(inp,inp+tot,cmp1);
    for(int i = cnt = 0; i < maxn; ++i)
        while(i == inp[cnt].k) inp[cnt++].ans = dp[i];
    sort(inp,inp+tot,cmp2);
    for(int i = 0; i < tot; ++i)
        printf("%d
",inp[i].ans);
    return 0;
}