HDU6130-Kolakoski

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本文介绍了一种特殊的数列——Kolakoski序列，该序列仅由1和2组成，首项为1。文章通过模拟的方式，探讨了如何生成这一序列，并提供了计算序列第n项的代码实现。

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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

n 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 integer

n(1≤n≤107) .

Output

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

Sample Input

Sample Output

1
2

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 long

int 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;
}