HDU6130-Kolakoski

Kolakoski

                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                                Total Submission(s): 677    Accepted Submission(s): 342

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

 

题意：给你一个序列，这个序列只包括1和2，并且第一项为1，同时如果把相邻且相同的项看成一组，计算每一组项的个数则能够得到这个序列本身

解题思路：模拟+打表即可

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>#include <string>#include <cmath>#include <algorithm>#include <vector>#include <bitset>#include <stack>#include <queue>#include <unordered_map>#include <functional>using namespace std;const int INF=0x3f3f3f3f;#define LL long longint a[10000009];void init(){    a[1]=1,a[2]=2;    int k=2,now=2;    for(int i=2;k<10000008;i++)    {        for(int j=k;j<k+a[i];j++) a[j]=now;        k=k+a[i];        if(now==2) now=1;        else now=2;    }}int main(){    init();    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        printf("%d\n",a[n]);    }    return 0;}