Adjacent Bit Counts
4557 Adjacent Bit Counts
For a string of n bits x 1 , x 2 , x 3 ,..., x n , 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 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
题意：给你一个长为n的串，且仅有0和1组成，x 1 ∗ x 2 + x 2 ∗ x 3 + x 3 ∗ x 4 + . . . + x n−1 ∗ x n 这样计算的值等于k，问该字符串有多少种这样的组合？
分析：用动态规划的思想，假如dp[i][j][k]表示组合数,i表示字串长度，j表示该串价值，k表示最后一个字符只能为1和0。
dp[i][j][0],有i个字符，价值为j,最后一位为0，这样的组合数等于长度为i-1，价值为j，最后一位为0的组合数加上长度为i-1，价值为j，最后一位为1的组合数，即dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1]。
dp[i][j][1],有i个字符，价值为j,最后一位为1，这样的组合数等于长度为i-1，价值为j，最后一位为0的组合数加上长度为i-1，价值为j-1,最后一位为1的组合数，即dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j-1][1]。
总结下来状态转移方程即是：
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]
再说说初始化
长度为1的时候，没有其他的数和它相乘，所以价值只能为0
dp[1][0][0]=1;长度为1价值为0且最后一位为0的组合数只有1种，比如0
dp[1][0][1]=1;长度为1价值为0且最后一位为1的组合数只有1种，比如1
看了别人题解，都说这是很简单的动规，佩服佩服
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 int dp[105][105][2];//dp定义为全局变量 5 void find_(){ 6 dp[1][0][0]=dp[1][0][1]=1;//初始化 7 for(int i=2;i<105;i++){//每一个长度时，都需要初始化一遍 8 dp[i][0][0]=dp[i-1][0][1]+dp[i-1][0][0]; 9 dp[i][0][1]=dp[i-1][0][0]; 10 for(int j=1;j<i;j++){//长度为i时，价值最多为i-1 11 dp[i][j][0]=dp[i-1][j][1]+dp[i-1][j][0]; 12 dp[i][j][1]=dp[i-1][j][0]+dp[i-1][j-1][1]; 13 } 14 } 15 } 16 int main() { 17 int p; 18 find_(); 19 scanf("%d",&p); 20 while(p--) { 21 int n,a,b; 22 scanf("%d %d %d",&n,&a,&b); 23 printf("%d %d ",n,dp[a][b][0]+dp[a][b][1]); 24 } 25 return 0; 26 } 27 28 CUTECode