POJ 2081
Recaman's Sequence
Time Limit: 3000MS | Memory Limit: 60000K | |
Total Submissions: 18575 | Accepted: 7751 |
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
7 10000 -1
Sample Output
20 18658
#include <stdio.h> #include <string.h> const int N=500010; int ch[N]; bool vis[N*10];//状态数组一定要开得比数字数组大,否则测试时会RE void init() { int i,j; memset(vis,false,sizeof(vis)); memset(ch,0,sizeof(ch)); vis[0]=vis[1]=vis[3]=true; ch[0]=0;ch[1]=1; for(i=2;i<N;i++) { ch[i]=ch[i-1]-i; if(ch[i]<1||vis[ch[i]]) ch[i]=ch[i-1]+i; vis[ch[i]]=true; } return ; } int main() { int i,j,k; init(); while(scanf("%d",&k),k!=-1) printf("%d\n",ch[k]); return 0; }
作者:火星十一郎
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