01-K Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 728    Accepted Submission(s): 209

Problem Description

Consider a code string S of N symbols, each symbol can only be 0 or 1. In any consecutive substring of S, the number of 0's differs from the number of 1's by at most K. How many such valid code strings S are there? 

For example, when N is 4, and K is 3, there are two invalid codes: 0000 and 1111. 

 

Input

The input consists of several test cases. For each case, there are two positive integers N and K in a line.

N is in the range of [1, 62].

K is in the range of [2, 5].

 

Output

Output the number of valid code strings in a line for each case.

 

Sample Input

 

1 2

4 3

 

Sample Output

2

14

//有一个长为 i 的 0-1 组成的序列，问任意连续长的子序列 0 1 数量差都不超过 m 的情况多少种

// dp[i][j][k] 意思是长为 i 的序列中 0 比 1 最多多 j 个，且长为 i 的序列中 1 比 0 最多多 k 个这样的序列的个数

所以 j，k最小为 0

dp [i][j][k] = dp[i-1][max(0,j-1)][k+1] // 填 1

                +dp[i-1][j+1][max(0,k-1)]  // 填 0

#include <bits/stdc++.h>
using namespace std;
#define LL long long
 
int n,m;
LL dp[105][10][10]; //长为 i 的序列，0比1最多多 j 个，且1比0最多多 k 个
 
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=1;
        for (int i=0;i<=n;i++)
        {
            for (int j=0;j<=m;j++)
            {
                for (int k=0;k<=m;k++)
                {
                    dp[i+1][j+1][max(k-1,0)]+=dp[i][j][k];  // 下一个数是 0
                    dp[i+1][max(j-1,0)][k+1]+=dp[i][j][k];  // 下一个数是 1
                }
            }
        }
        LL ans = 0;
        for (int i=0;i<=m;i++)
            for (int j=0;j<=m;j++)
                ans += dp[n][i][j];
        printf("%lld
",ans);
    }
    return 0;
}