HDU - 6130 Kolakoski （打表）

题意：由1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1，……合并可得1,22,11,2,1,22,1,22,11,2,11,22,1，再由每个数的位数可得新序列，推出新序列第n项。

分析：新序列与原序列相同，按题意打表即可。

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define lowbit(x) (x & (-x))

const double eps = 1e-8;

inline int dcmp(double a, double b){

    if(fabs(a - b) < eps) return 0;

    return a > b ? 1 : -1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 10000 + 10;

const int MAXT = 10000 + 10;

using namespace std;

vector<int> v;

int main(){

    v.push_back(1);

    v.push_back(2);

    v.push_back(2);

    v.push_back(1);

    v.push_back(1);

    int cur = 3;

    while(1){

        int cnt = v[cur];

        int x;

        if(v.size() & 1){

            x = 2;

        }

        else{

            x = 1;

        }

        for(int i = 0; i < cnt; ++i){

            v.push_back(x);

        }

        if(v.size() > 10000000) break;

    }

    int T;

    scanf("%d", &T);

    while(T--){

        int n;

        scanf("%d", &n);

        printf("%d\n", v[n - 1]);

    }

    return 0;

}