BNU4286——Adjacent Bit Counts——————【dp】

Adjacent Bit Counts

Time Limit: 1000ms
Memory Limit: 65536KB
64-bit integer IO format:  %lld      Java class name: Main
Type: 
None
 
 
For a string of  bits  x 1 x 2 ,  x 3 , …,  x n,B the  adjacent bit count of the string ( AdjBC(x)) is given by
x 1 *x 2  + x 2 *x 3  + x 3 *x 4  + … + 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  and  and returns the number of bit strings  of  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 ≤ ≤ 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 and 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


解题思路:用dp[i][j][k]来定义状态。i表示当前数字是第i位,j表示达到j值,k代表末尾是0还是1.由于当末尾为0对新加的一位没有要求,即j值不会变动,所以当dp[i][j][0]时转移方程为dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1]。而当末尾为1时如果新加一位的值为1,会影响j的值。所以当dp[i][j][1]时dp[i][j][1]=dp[i-1][j-1][1]+dp[i-1][j][0]。



#include<bits/stdc++.h>

using namespace std;

int dp[500][500][2];

void DP(int n){

    memset(dp,0,sizeof(dp));

    dp[1][0][0]=1;

    dp[1][0][1]=1;

    for(int i=2;i<n;i++){

        for(int j=0;j<i;j++){

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

            if(j){

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

            }else{

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

            }



        }

    }

}

int main(){

    int n;

    scanf("%d",&n);

    DP(110);

    while(n--){

        int t,ta,tb;

        scanf("%d%d%d",&t,&ta,&tb);

        cout<<t<<" "<<dp[ta][tb][0]+dp[ta][tb][1]<<endl;



    }

    return 0;

}

  

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