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Recaman's Sequence_递推
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.
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.
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
710000-1
Sample Output
2018658
【题意】第m数是根据第m-1数推出来。如果a[m-1]-m>0,并且a[m-1]-m在前面的序列中没有出现过那么a[m] = a[m-1]-m否则a[m] = a[m-1]+m
#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int N=10000007;int a[500005];bool ha[N];int main(){ memset(a,0,sizeof(a)); memset(ha,false,sizeof(ha)); for(int i=1;i<=500000;i++) { if(a[i-1]-i>0&&!ha[a[i-1]-i]) { a[i]=a[i-1]-i; } else a[i]=a[i-1]+i; ha[a[i]]=true; } int k; while(~scanf("%d",&k)) { if(k<0) break; printf("%d\n",a[k]); } return 0;}
Recaman's Sequence_递推
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