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
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
Tags: dynamic programming ( hide tags for unsolved problems
Difficulty: 255
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