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 = ;
 int dp[maxn];
 bool vis[];
 struct node {
     int k,ans,o;
 };
 node inp[];
 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 = ,tmp,cnt;
     for(int i = ; i < maxn; ++i) {
         dp[i] = dp[i-]-i;
         if(dp[i] <=  || vis[dp[i]])
             dp[i] = dp[i-]+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 = ; i < maxn; ++i)
         while(i == inp[cnt].k) inp[cnt++].ans = dp[i];
     sort(inp,inp+tot,cmp2);
     for(int i = ; i < tot; ++i)
         printf("%d\n",inp[i].ans);
     return ;
 }