Adjacent Bit Counts(uvalive)

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 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

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
题解：求一个长度为p，构造一个价值为n的字符串的方法数；

首先如果我们要构造一个价值为7的字符串，若分为两部分的话，我们有1+6，2+5，3+4等3种方案，对于1+6，我们又可以分为一个子问题，将6分为两部分，以此类推；

这样就满足一个最优子结构；这样我们把原问题可以分为若干子问题，我们用的dp[i][j]表示长度为i-1，价值为j的字符串的方法数，由于最后一位0与1两种情况，我们可以定义3维数组；

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 <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long ll;
const int MAXN=1e3+10;
int m,n;
int dp[MAXN][MAXN][2];
using namespace std;
int  main()
{
    cin>>m;
    int p,k,u;
   while(m--)
   {
       cin>>p>>k>>u;
       memset(dp,0,sizeof(dp));
       dp[1][0][1]=1;
       dp[1][0][0]=1;
       for(int i=2;i<=k;i++)
       {
           for(int j=0;j<=u;j++)
           {
               for(int t=0;t<2;t++)
               {
                   if(t==0)
                   {
                       dp[i][j][t]=dp[i-1][j][0]+dp[i-1][j][1];
                   }
                   else
                   {
                       dp[i][j][t]=dp[i-1][j][0]+dp[i-1][j-1][1];
                   }
               }
           }
       }
       cout<<p<<" "<<dp[k][u][0]+dp[k][u][1]<<endl;
   }
}

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