- 描述
-
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
- 输入
- 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
- 输出
- For each test case, output a line with the number of n-bit strings with adjacent bit count equal to p.
- 样例输入
-
2 5 2 20 8
- 样例输出
-
6 63426
讲解：看了半天没有看出来，其实就是一个dp问题；看下代码：1 #include<algorithm> 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 using namespace std; 6 long long dp[105][105][2]; 7 void fun() 8 { int i,j; 9 memset(dp,0,sizeof(dp)); 10 dp[1][0][0]=1;dp[1][0][1]=1; 11 for(i=2;i<=100;i++) 12 { 13 dp[i][0][0]=dp[i-1][0][1]+dp[i-1][0][0]; 14 dp[i][0][1]=dp[i-1][0][0]; 15 dp[i][i-1][1]=1; 16 } 17 for(j=1;j<=100;j++) 18 for(i=j+2;i<=100;i++) 19 { 20 dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1]; 21 dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j-1][1]; 22 } 23 } 24 int main() 25 { 26 fun(); 27 int t,m,n; 28 cin>>t; 29 while(t--) 30 { 31 cin>>m>>n; 32 cout<<dp[m][n][1]+dp[m][n][0]<<endl; 33 } 34 return 0; 35 }