hdu 6130 Kolakoski（多校联赛）_k - kolakoski hdu

本文介绍了一种特殊的数列——Kolakoski数列，通过递推方式生成序列，并提供了一段C++代码实现其计算过程。文章探讨了数列构造规则及其实现方法。

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Kolakoski

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

2
1
2

Sample Output

1
2

水题一道但是学到了一个新的高深而难懂数列

ac代码：

#include<bits/stdc++.h>
#define inf 10000000
using namespace std;
int a[10000000+5];
int ke()
{
    a[1]=1,a[2]=2,a[3]=2;
    int sum=3;
    for(int i=3;i<=inf;i++)
    {
        if(a[i]==2)
        {
            if(a[sum]==2)
            {
                a[sum+1]=a[sum+2]=1;
            }
            if(a[sum]==1)
            {
                a[sum+1]=a[sum+2]=2;
            }
        }
        if(a[i]==1)
        {
            if(a[sum]==2)
            {
                a[sum+1]=1;
            }
            if(a[sum]==1)
            {
                a[sum+1]=2;
            }
        }
        sum=sum+a[i];
        if(sum>=inf)
            break;
    }
}
int main()
{
    ke();
    int t,n;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d",&n);
            printf("%d\n",a[n]);
        }
    }
    return 0;
}