HDU-Kolakoski

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最新推荐文章于 2019-09-25 17:33:22 发布

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分类专栏： ACM_数学 ACM_杂

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本文介绍了一种特殊的数列——Kolakoski序列，并提供了一个算法实现方案，用于计算该序列中任意位置的元素值。通过递增生成序列的方式，确保了序列的独特性和准确性。

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Kolakoski

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1287 Accepted Submission(s): 884

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

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
const int maxn = 1e7+10;
using namespace std;

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)
    {
        {
            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;
}