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HUNAN UNIVERSITY ACM/ICPC Judge Online_Problem 10010_Recaman's Sequence

2008-06-21 11:55 567 查看

Recaman's Sequence

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 549, Accepted users: 471

Problem 10010 : No special judgement

Problem description

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

Problem Source

Shanghai-P 2004

CODE

/*

Name:  10010_Recaman's Sequence

Copyright:  yangchun's

Author:  yangchun

Date:  20-06-08 22:27

Description:  HUNAN UNIVERSITY ACM/ICPC Judge Online_Problem 10010_Recaman's Sequence

*/

#include <algorithm>

#include <iostream>

using namespace std;

long sequence[500000];

int _sequence[5000000];

int main()

{

// freopen("in.txt","r",stdin);

//freopen("out.txt","w",stdout);

sequence[0] = 0;

for(long m=1; m<500000; m++)

{

sequence[m] = sequence[m-1] - m;

if(sequence[m] <=0 || _sequence[sequence[m]] == 1)

sequence[m] = sequence[m-1] + m;

_sequence[sequence[m]] = 1;

}

long n=0;

while(scanf("%d",&n),n!=-1)

printf("%ld/n",sequence
);

return 0;

}

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