URAL1081

Consider all the sequences with length (0 <  N < 44), containing only the elements 0 and 1, and no two ones are adjacent (110 is not a valid sequence of length 3, 0101 is a valid sequence of length 4). Write a program which finds the sequence, which is on K-th place (0 <  K < 10 9) in the lexicographically sorted in ascending order collection of the described sequences.

Input

The first line of input contains two positive integers N and K.

Output

Write the found sequence or −1 if the number K is larger then the number of valid sequences.

Example

input	output
3 1	000

思路：先判断输入的K是否合法（一开始用组合里的隔板法判断长度，后来发现是个斐波那契数列，）

• 隔板法，用0将1隔开，考虑0的数量和1的数量就可以得到该长度下01字符串的种数
• 字符串的长度满足斐波那契数列，f(i)=f(i-1)+f(i-2)，只要在长为i-2和i-1的字符串前加1或者0

然后要构造字典序第K小的序列，我们要根据K和f(i)的关系判断这一位是0还是1

 

#include<cstdio>
#include<string.h>
using namespace std;

long long fac[15];
long long num[50];

//长为N的字符串的可能数构成斐波那契数列
void init()
{
    num[0] = 0;
    num[1] = 2; num[2] = 3;
    for (int i = 3; i <= 44; i++)
        num[i] = num[i - 1] + num[i - 2];
}

////寻找长为N的字符串有几种可能性
//bool check(int N, int K)
//{
//    if (num[N] != -1)
//    {
//        if (num[N] > K)
//            return true;
//        else
//            return false;
//    }
//
//    long long res = 0;
//    for (int i = 1; i <= N; i++)
//    {
//        if (N - (i - 1) < i)
//            break;
//        res += fac[N - (i - 1)] / fac[i] / fac[N - (i - 1) - i];
//    }
//    res++;
//    num[N] = res;
//    if (res >= K)
//        return true;
//    else
//        return false;
//}

void findk(int N, int K)
{
    int res[50];    //结果
    for (int i = 1; i <= N; i++)
        res[i] = 0;

    int tail[50];   //末尾的个数
    for (int i = 0; i < 50; i++)
        tail[i] = 0;

    int p = 0;  //可能要改变的末尾的p个数
    for (int i = 1; i <= N; i++)
        if (K <= num[i])
        {
             p = i;
             break;
        }


    int pos = 0;

    for (int i = p; i >= 1; i--)
    {
        if (K == 1) //最后一位为0
        {
            for (; pos < p; pos++)
                    tail[pos] = 0;
            break;
        }
        //K=num[i],则找字典序最大的情况
        if (K == num[i])
        {
            for (; pos < p; pos++)
            {
                if (pos == 0)
                    tail[pos] = 1;
                else if (tail[pos - 1] == 0)
                    tail[pos] = 1;
                else if (tail[pos - 1] == 1)
                    tail[pos] = 0;
            }
            break;
        }
        else if (K < num[i] && K > num[i - 1])
        {
            tail[pos++] = 1;
            K = K - num[i - 1];
        }
        else if (K < num[i] && K <= num[i - 1])
            tail[pos++] = 0;


    for (int i = N - p + 1, j = 0; i <= N; i++)
        res[i] = tail[j++];

    for (int i = 1; i <= N; i++)
        printf("%d", res[i]);
    printf("\n");
}

int main()
{
    memset(num, -1, sizeof(num));
    fac[0] = 1;
    for (int i = 1; i <= 15; i++)
        fac[i] = i * fac[i - 1];

    int N, K;
    init();

    while (scanf("%d %d", &N, &K) != EOF)
    {
        if (K == 1)
        {
            for (int i = 1; i <= N; i++)
                printf("0");
            printf("\n");
        }
        else if (num[N] >= K)
            findk(N, K);
        else
            printf("-1\n");
    }

    return 0;
}