【nyoj715】Adjacent Bit Counts

Description

For a string of n bits x1, x2, x3, …, xn,  the adjacent bit count of the string  is given by     fun(x) = x1*x2 + x2*x3 + x3*x 4 + … + xn-1*x n
which counts the number of times a 1 bit is adjacent to another 1 bit. For example: 
     Fun(011101101) = 3
     Fun(111101101) = 4
     Fun (010101010) = 0
Write a program which takes as input integers n and p and returns the number of bit strings x of n bits (out of 2ⁿ) that satisfy  Fun(x) = p.

For example, for 5 bit strings, there are 6 ways of getting fun(x) = 2:
11100, 01110, 00111, 10111, 11101, 11011

Input

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 10  Each case is a single line that contains  a decimal integer giving the number (n) of bits in the bit strings, followed by a single space, followed by a decimal integer (p) giving the desired adjacent bit count. 1 ≤ n , p ≤ 100

Output

For each test case, output a line with the number of n-bit strings with adjacent bit count equal to p.

Sample Input

2

5 2

20 8

Sample Output

6

63426

题意：给n,p，代表01串的长度是n，运算后是p，运算规则为x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn，求这样的01串共有几个。

思路：dp[i][j][k]代表01串长度为i，运算结果是j，最后一位数是k。所以最终答案是dp[n][p][0]+dp[n][p][1].

转移方程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].根据n p范围，所以有：

i：2~n

j：1~p

循环想好后注意初始化，当i=1时，考虑到dp[1][1][0]=0; dp[1][1][1]=0; dp[1][0][1]=1;

但是还存在一组特殊情况：j-1可能等于0，dp[1][0][1]=1,dp[2][0][1]=1,dp[3][0][1]=2,dp[4][0][1]=3,dp[5][0][1]=5........所以dp[i][0][1]=dp[i-1][0][1]+dp[i-2][0][1]。

代码：

#include<bits/stdc++.h>using namespace std;int dp[105][105][2];int main(){    int n,m,i,j,t;    dp[1][1][0]=0;    dp[1][1][1]=0;    dp[1][0][1]=1;    dp[2][0][1]=1;    for(i=3; i<=100; i++)        dp[i][0][1]=dp[i-1][0][1]+dp[i-2][0][1];    for(i=2; i<=100; i++)        for(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][0]+dp[i-1][j-1][1];        }    cin>>t;    while(t--)    {        scanf("%d%d",&n,&m);        printf("%d\n",dp[n][m][0]+dp[n][m][1]);    }}