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