hdu 6130 Kolakoski
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Kolakoski
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 623 Accepted Submission(s): 307
Problem Description
This is Kolakosiki sequence: 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1…… . This sequence consists of 1 and 2 , and its first term equals 1 . Besides, if you see adjacent and equal terms as one group, you will get 1,22,11,2,1,22,1,22,11,2,11,22,1…… . Count number of terms in every group, you will get the sequence itself. Now, the sequence can be uniquely determined. Please tell HazelFan itsn th element.
Input
The first line contains a positive integer T(1≤T≤5) , denoting the number of test cases.
For each test case:
A single line contains a positive integern(1≤n≤107) .
For each test case:
A single line contains a positive integer
Output
For each test case:
A single line contains a nonnegative integer, denoting the answer.
A single line contains a nonnegative integer, denoting the answer.
Sample Input
212
Sample Output
12
Kolakoski序列见百度百科https://baike.baidu.com/item/Kolakoski%E5%BA%8F%E5%88%97/16683856?fr=aladdin
枚举就行
代码
//https://baike.baidu.com/item/Kolakoski%E5%BA%8F%E5%88%97/16683856?fr=aladdin#include <iostream>#include <cstdio>#include <cmath>#include <cstring> using namespace std;const int maxn = 1e7+10;int a[maxn];int main(){ a[1] = 1,a[2] = 2,a[3] = 2; int cnt = 3; for(int i = 3;i<maxn&&cnt<maxn;++i){ for(int j = 1;j<=a[i];j++){ if(a[cnt] == 2){ a[cnt+j] = 1; } else a[cnt+j] = 2; } cnt += a[i]; } int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); printf("%d\n",a[n]); } return 0;}
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