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, …, xn, the adjacent bit count of the string (AdjBC(x)) is given by

x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn

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

10
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

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;
}