第六届程序设计大赛 Adjacent Bit Counts（01串DP）

Adjacent Bit Counts

时间限制: 1 Sec  内存限制: 65535 MB

题目描述

For a string of n bits x1, x2, x3, …, xn,  the adjacent bit count of the string  is given by     fun(x) = x₁*x₂ + x₂*x₃ + x₃*x ₄ + … + x_n-1*x _n
which counts the number of times a 1 bit is adjacent to another 1 bit. For example:  

     Fun(011101101) = 3

     Fun(111101101) = 4

     Fun (010101010) = 0

Write a program which takes as input integers n and p and returns the number of bit strings x of n bits (out of 2ⁿ) that satisfy  Fun(x) = p.

 

For example, for 5 bit strings, there are 6 ways of getting fun(x) = 2:

11100, 01110, 00111, 10111, 11101, 11011

输入

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10 Each case is a single line that contains a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (p) giving the desired adjacent bit count. 1 ≤ n , p ≤ 100

输出

For each test case, output a line with the number of n-bit strings with adjacent bit count equal to p.

样例输入

25 220 8

样例输出

663426

提示

  河南省第六届ACM程序设计大赛

题意：

一种只有0、1两种元素的串，每个串有一个权值 x₁*x₂ + x₂*x₃ + x₃*x₄ + … + x_n-1*x_n。给出你某个串的长度 n，求出其权值为 k 时的方案种数。

解题思路：

这么水的DP都不会，哎。好像见过几道关于01串的dp问题了，下次遇到可以往这方面想想。

分析：对于长度为 i 的串，假设它的权值为 j，则长度为 i+1 的串的权值只可能为 j 或 j+1，且仅与末位元素和新添加元素有关。

令 dp[i][j][k] 表示长度为 i 的串、权值为 j 、末位为 k (0 or 1) 的方案种数。

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]。

#include <stdio.h>   
long long  dp[102][102][2]; 
int main() 
{ 
    int z,ca,n,k; 
    scanf("%d",&z); 
    dp[1][0][0] = dp[1][0][1] = 1; 
    for(int i=2;i<=100;i++) 
    { 
        dp[i][0][0] = dp[i-1][0][0] + dp[i-1][0][1]; 
        dp[i][0][1] = dp[i-1][0][0]; 
        for(int j=1;j<i;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]; 
        } 
    } 
    while(z--) 
    { 
        scanf("%d%d",&n,&k); 
        printf("%lld\n",dp[n][k][0]+dp[n][k][1]); 
    } 
    return 0; 
}