简单DP---Adjacent Bit Counts

原创于 2018-10-17 20:29:38 发布 · 219 阅读

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本文介绍了一种使用动态规划解决的问题，旨在计算长度为n的二进制字符串中，相邻1位的数量等于给定值k的不同组合数。通过递推公式，文章详细解释了如何高效地解决这个问题，并提供了完整的AC代码实现。

Description

For a string of n bits x1, x2, x3,…, xn, the adjacent bit count of the string (AdjBC(x)) is given by

$x_{1}*x_{2}+x_{2}*x_{3}+x_{3}*x_{4}...x_{n-1}*x_{n}$

which counts the number of times a 1 bit is adjacent to another 1 bit. For example:

AdjBC(011101101) = 3

AdjBC(111101101) = 4

AdjBC(010101010) = 0

Write a program which takes as input integers n and k and returns the number of bit strings x of n bits (out of $2^{n}$ ) that satisfy AdjBC(x) = k. For example, for 5 bit strings, there are 6 ways of getting AdjBC(x) = 2:

11100, 01110, 00111, 10111, 11101, 11011

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (k) giving the desired adjacent bit count. The number of bits (n) will not be greater than 100 and the parameters n and k will be chosen so that the result will fit in a signed 32-bit integer.

Output

For each data set there is one line of output. It contains the data set number followed by a single space, followed by the number of n-bit strings with adjacent bit count equal to k.

Sample Input

1 5 2

2 20 8

3 30 17

4 40 24

5 50 37

6 60 52

7 70 59

8 80 73

9 90 84

10 100 90

Sample Output

1 6

2 63426

3 1861225

4 168212501

5 44874764

6 160916

7 22937308

8 99167

9 15476

10 23076518

题意

简单来说就是长度为n时按照所给公式得到的值为k的串的组成方式有几种

思路

每周组队赛抓的一道题，当时没做出来，后来才发现是一个简单的DP，不难想到长度为n的串可以由长度为n-1的串得到，只需要分别考虑末尾是以1还是0结尾即可，若长度为n时末尾为0，则当n-1的串添加0时值是不变的，可得到它为n-1长度时值已经为k，同理可以推出末尾为1的情况，即状态转移方程为：

dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1];

dp[i][j][1] = dp[i-1][j][0] + dp[i-1][j-1][1];

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 1e3+10;

int dp[MAXN][MAXN][2];//三维分别为长度，值，0结尾还是1结尾

int main()
{
    int T;
    scanf("%d",&T);
    int x,n,k;
    while(T--){
        scanf("%d%d%d",&x,&n,&k);
        memset(dp,0,sizeof(dp));
        dp[1][0][1] = 1;
        dp[1][0][0] = 1;
        for(int i=2;i<=n;i++){
            for(int j=0;j<=k;j++){
                dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1];
                dp[i][j][1] = dp[i-1][j][0] + dp[i-1][j-1][1];
            }
        }
        printf("%d ",x);
        printf("%d\n",dp[n][k][0]+dp[n][k][1]);
    }
    return 0;
}