HDU6130-Kolakoski

Kolakoski

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 386 Accepted Submission(s): 184

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

Source
2017 Multi-University Training Contest - Team 7

题目大意：求Kolakoski sequence的第n项。
解题思路：在OEIS和Wikipedia上查了一下，发现递推就好，设置当前指针和尾指针，预处理就好， 107 10^7的数组，用布尔类型开~

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#define lson u<<1
#define rson u<<1|1
using namespace std;
const int maxn = 2e5 + 19, INF = 0x7fffffff;
const int MAXN=1e7+10;
bool a[MAXN];int main() {int T;scanf("%d",&T);a[1]=false;a[2]=true;a[3]=true;int tail=3;for(int i=3;;i++){if(tail>=1e7) break;if(a[i]==false){if(a[tail]==false) {tail++;a[tail]=true;}else  {tail++;a[tail]=false;}}else{if(a[tail]==false) {tail++;a[tail]=true;tail++;a[tail]=true;}else  {tail++;a[tail]=false;tail++;a[tail]=false;}}}while(T--){int n;scanf("%d",&n);
//        for(int i=1;i<=20;i++)
//            cout<<a[i]<<" ";if(a[n]==true) printf("2\n");else printf("1\n");}return 0;
}

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