Ural binary lexicographic sequence (DP + DFS)

Http://acm.timus.ru/problem.aspx? Space = 1 & num = 1081

It is legal to have a binary sequence defined as not having two consecutive 1 occurrences. Given the length N of the sequence, find out what the K sequence is after the valid binary sequence is ordered by lexicographically.

Set DP [I] [0] and DP [I] [1] to indicate the numbers 0 and 1 on the I-th, respectively.

Then DP [I] [0] = DP [I-1] [0] + dp [I-1] [1]; DP [I] [1] = DP [I-1] [0], the table is found to be similar to the Fibonacci series.

When the k-th legal sequence is obtained, it can be found that when k> = DP [n-1], this bit is 1; otherwise, this bit is 0.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL __int64#define eps 1e-12#define PI acos(-1.0)using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 4010;int dp[50][2],tmp[50];void init(){memset(dp,0,sizeof(dp));dp[1][0] = dp[1][1] = 1;tmp[1] = 2;for(int i = 2; i < 44; i++){dp[i][0] += (dp[i-1][0] + dp[i-1][1]);dp[i][1] += dp[i-1][0];tmp[i] = dp[i][0] + dp[i][1];}}void dfs(int n, int k){if(n == 1){if(k == 1)printf("0");elseprintf("1");return;}if(k <= tmp[n-1]){printf("0");dfs(n-1,k);}else{printf("1");dfs(n-1,k-tmp[n-1]);}}int main(){init();int n,k;while(~scanf("%d %d",&n,&k)){if(k > tmp[n]){printf("-1\n");continue;}dfs(n,k);printf("\n");}return 0;}

Ural binary lexicographic sequence (DP + DFS)