大体题意:
定义一个合法的二进制序列为序列中没有两个1是相邻的,对于所有长度为n的合法序列按照字典序排序后(保留前导0),求第K大的串是多少?
思路:
我们先得求出一个n位的合法二进制的个数是多少!
f[1] = 2;
f[2] = 3
f[3] = 5
f[4] = 8
这几个很容易算出来,然后我们就很明显的发现规律了,这是一个斐波那契数列!
其实想一想确实这个样!
你构造第n位时(高位),要么填0,要么填1,填0的话,直接是f(n-1),填1的话,要填10,结果是f(n-2)
因此 f(n) = f(n-1) + f(n-2);
然后在考虑第K大!
f(n-2)的数 第一个数是1,f(n-1) 第一个数是0,显然f(n-1)字典序要小!
因此如果k <= f(n-1) ,那么这一位必然填0,继续递归!
相反如果k > f(n-1) 这一位必然是10,继续递归!
注意 k 不合法时,要输出-1,没看到wa了一次!
详细见代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> using namespace std; typedef long long ll; int n, k; ll f[50]; void init(){ f[0] = 1; f[1] = 2; for (int i = 2; i <= 45; ++i){ f[i] = f[i-1] + f[i-2]; } } string ans; void dfs(int c,int o){ if (c == 0){ return; } if (c == 1){ if (o == 1) { ans += "0"; return; } ans += "1"; return; } if (f[c-1] >= o){ ans += "0"; dfs(c-1,o); } else { ans += "10"; dfs(c-2,o-f[c-1]); } } int main(){ init(); while(scanf("%d %d",&n, &k)!=EOF){ ans = ""; if (k > f[n]){ puts("-1"); continue; } dfs(n,k); printf("%s\n",ans.c_str()); } return 0; }
1081. Binary Lexicographic Sequence
Time limit: 0.5 second
Memory limit: 64 MB
Memory limit: 64 MB
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.
Sample
input | output |
---|---|
3 1 |
000 |
Problem Author: Emil Kelevedzhiev
Problem Source: Winter Mathematical Festival Varna '2001 Informatics Tournament
Problem Source: Winter Mathematical Festival Varna '2001 Informatics Tournament
Tags:
dynamic programming
)
Difficulty: 255
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