HDU 6130 Kolakoski

Kolakoski

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 955    Accepted Submission(s): 562

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 its nth 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 integer n(1≤n≤107).

 

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

 

Sample Input

212

 

Sample Output

12

 

Source

2017 Multi-University Training Contest - Team 7

 

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找规律，看第二个数列，用逗号隔开的，是1212这种轮回，然后，慢慢的就墨迹出来规律了

My ugly code

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <iostream>#include <algorithm>#define ll long longusing namespace std;const int maxn=1e7+10;int t,n;int a[maxn]={0,1,2,2,1};void INIT(){    int k=3;    for(int i=3;i<=1e7 && k<=1e7;i++){        if(a[i] == 2){            if(a[k]==2){                a[k+1]=1;                a[k+2]=1;            }            else{                a[k+1]=2;                a[k+2]=2;            }            k=k+2;        }        else{            if(a[k]==1){                a[k+1]=2;            }            else{                a[k+1]=1;            }            k=k+1;        }    }}int main(){    INIT();    scanf("%d",&t);    while(t--){        scanf("%d",&n);        printf("%d\n",a[n]);    }    return 0;}