hdu--3284

Adjacent Bit Counts

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 378    Accepted Submission(s): 308

Problem Description

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

x₁*x₂ + x₂*x₃ + x₃*x₄ + … + 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 2ⁿ) 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

101 5 22 20 83 30 174 40 245 50 376 60 527 70 598 80 739 90 8410 100 90

 

Sample Output

1 62 634263 18612254 1682125015 448747646 1609167 229373088 991679 1547610 23076518

 

Source

2009 Greater New York Regional

解体思路:简单的dp,想了很久也没想到,感觉dp好强大.得好好花点功夫学习

代码如下:

#include<stdio.h>int dp[110][110][2];//d[i][j][k]表示长度为i的串,权值为j时,以k=(1 or 0)为尾时的方法数 int main(){int t,l,p,cas;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][1]+dp[i-1][j][0];    dp[i][j][1]=dp[i-1][j-1][1]+dp[i-1][j][0];    }}scanf("%d",&t);while(t--){scanf("%d%d%d",&cas,&l,&p);    printf("%d %d\n",cas,dp[l][p][0]+dp[l][p][1]);}return 0;}