题意：给出一个描述自身的数列，求出第n项

思路：看了很久题目才看懂..每个值其实是描述一个分组中的个数，把两个数列对照一下就可以了，那么一个指针扫，同时向尾部加数，构造个数组就行了。其实很水..

 

/** @Date    : 2017-08-15 12:13:59  * @FileName: 1011.cpp  * @Platform: Windows  * @Author  : Lweleth (SoungEarlf@gmail.com)  * @Link    : https://github.com/  * @Version : $Id$  */#include 
    
     #define LL long long#define PII pair
     
      #define MP(x, y) make_pair((x),(y))#define fi first#define se second#define PB(x) push_back((x))#define MMG(x) memset((x), -1,sizeof(x))#define MMF(x) memset((x),0,sizeof(x))#define MMI(x) memset((x), INF, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int N = 1e7+20;const double eps = 1e-8;int a[N] = {0, 1,2,2,1,1,2,1,2,2,1,2,2,1,   1, 2, 1, 1, 2, 2, 1};int main(){	int cnt = 21;	for(int i = 14; cnt <= 10000000; i++) {		if(a[i] == 0)			break;		if(a[i] == 1) {			a[cnt] = (3 - a[cnt - 1]);			cnt++;		}		else if(a[i] == 2) {			a[cnt] = (3 - a[cnt - 1]);			a[cnt + 1] = (3 - a[cnt - 1]);			cnt += 2;		}	}	int T;	cin >> T;	while(T--)	{		int n;		scanf("%d", &n);		printf("%d\n", a[n]);	}    return 0;}