POJ-2081 Recaman's Sequence
2016-01-07 19:59
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Recaman’s Sequence
Time Limit: 3000MS Memory Limit: 60000K
Total Submissions: 22392 Accepted: 9614
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
Time Limit: 3000MS Memory Limit: 60000K
Total Submissions: 22392 Accepted: 9614
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
#include <iostream> #include <string.h> #include <math.h> #include <algorithm> #include <stdlib.h> using namespace std; #define MAX 500000 long long int a[MAX+5]; bool tag[10000000]; void fun() { a[0]=0; tag[0]=1; for(int i=1;i<=MAX;i++) { a[i]=a[i-1]-i; if(a[i]<0||tag[a[i]]==1) a[i]=a[i-1]+i; tag[a[i]]=1; } } int main() { int n; memset(tag,0,sizeof(tag)); fun(); while(scanf("%d",&n)!=EOF) { if(n==-1) break; printf("%d\n",a ); } }
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