POJ 3786 dp-递推 Adjacent Bit Counts *

Adjacent Bit Counts

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 2?) 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

Greater New York Regional 2009

1. 由0,1组成的长度为n的数列x1,x2,x3,x4.....xn,定义一个操作为
2. AdjBC(x) = x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn 。
3. 输入两个数n和m,求长度为n的数列,经过上述操作,最后结果为m共有多少种
4. 思路 : DP，递推
5. d[i][j][0]：表示前i项组成和为j且第i项为0共有多少种
6. d[i][j][1]：表示前i项组成和为j且第i项为1共有多少种
7. 状态转移方程：
8. d[i][j][1] = d[i-1][j][0] + d[i-1][j-1][1];
9. d[i][j][0] = d[i-1][j][1] + d[i-1][j][0]

   #include <stdio.h>
   #include <math.h>
   #include <iostream>
   #include <string.h>
   #include <algorithm>
   #include <queue>
   #define maxn 105
   using namespace std;
   int main()
   {
       int t,n,m;
       int dp[maxn][maxn][2];
       dp[1][0][0] = dp[1][0][1] = 1;
       for(int i=2;i<maxn;i++){
           dp[i][0][0] = dp[i-1][0][1] + dp[i-1][0][0];
           dp[i][0][1] = dp[i-1][0][0];
       }
       for(int i=2;i<maxn;i++){
           for(int j=1;j<maxn;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];
           }
       }
       int T;
       cin >> T;
       while(T--){
           cin >> t >> n >> m;
           cout << t << " " << dp[n][m][0]+dp[n][m][1] << endl;
       }
       return 0;
   }