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。。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 #include <stack>
 8 #include <set>
 9 #include <map>
10 #include <queue>
11 #include <ctime>
12 #define LL long long
13 #define INF 0x3f3f3f3f
14 #define pii pair<int,int>
15 
16 using namespace std;
17 const int maxn = 500010;
18 int dp[maxn];
19 bool vis[5000000];
20 struct node {
21     int k,ans,o;
22 };
23 node inp[1000000];
24 bool cmp1(const node &x,const node &y) {
25     return x.k < y.k;
26 }
27 bool cmp2(const node &x,const node &y) {
28     return x.o < y.o;
29 }
30 int main() {
31     int tot = 0,tmp,cnt;
32     for(int i = 1; i < maxn; ++i) {
33         dp[i] = dp[i-1]-i;
34         if(dp[i] <= 0 || vis[dp[i]])
35             dp[i] = dp[i-1]+i;
36         vis[dp[i]] = true;
37     }
38     while(~scanf("%d",&tmp)&&(~tmp)) {
39         inp[tot].k = tmp;
40         inp[tot].o = tot++;
41     }
42     sort(inp,inp+tot,cmp1);
43     for(int i = cnt = 0; i < maxn; ++i)
44         while(i == inp[cnt].k) inp[cnt++].ans = dp[i];
45     sort(inp,inp+tot,cmp2);
46     for(int i = 0; i < tot; ++i)
47         printf("%d\n",inp[i].ans);
48     return 0;
49 }
View Code

 

posted @ 2015-01-20 16:25  狂徒归来  阅读(297)  评论(2)  编辑  收藏  举报