poj 2081 Recaman's Sequence
Recaman's Sequence
| Time Limit: 3000MS | Memory Limit: 60000K | |
| Total Submissions: 22529 | Accepted: 9678 |
Description
The Recaman's sequence is defined by a0 = 0 ; for m > 0, a
m = a
m−1 − m if the rsulting a
m is positive and not already in the sequence, otherwise a
m = a
m−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 a k.
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 a k.
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 a
k to the output.
Sample Input
7 10000 -1
Sample Output
20 18658
Source
Shanghai 2004 Preliminary
#include<stdio.h>
#include<string.h>
int res[500005];
bool vis[3000005]; //res[i]的结果会很大，要开大一点vis[]
int main(){
int n;
memset(vis,true,sizeof(vis));
res[0]=0;
for(int i=1;i<=500000;i++){
int x=res[i-1]-i;
if(x>0&&vis[x]){ //判断当前值是否是正数并且保证不在res中
res[i]=res[i-1]-i;
vis[x]=false;
}
else {
res[i]=res[i-1]+i;
vis[res[i]]=false;
}
}
while(scanf("%d",&n)){
if(n<0) break;
printf("%d\n",res[n]);
}
return 0;
}