Adjacent Bit Counts
时间限制：1000 ms | 内存限制：65535 KB难度：4- 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 - 来源
- 第六届河南省程序设计大赛
- 上传者
- ACM_赵铭浩
思路：http://blog.csdn.net/dgq8211/article/details/8041473
题意：
一种只有0、1两种元素的串，每个串有一个权值 x1*x2 + x2*x3 + x3*x4 + … + xn-1*xn。给出你某个串的长度 n，求出其权值为 k 时的方案种数。
解题思路：
这么水的DP都不会，哎。好像见过几道关于01串的dp问题了，下次遇到可以往这方面想想。
分析：对于长度为 i 的串，假设它的权值为 j，则长度为 i+1 的串的权值只可能为 j 或 j+1，且仅与末位元素和新添加元素有关。
令 dp[i][j][k] 表示长度为 i 的串、权值为 j 、末位为 k (0 or 1) 的方案种数。
状态转移方程为 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]。
ac代码
#include<stdio.h>
#include<string.h>
int dp[105][105][2];
int main()
{
int t,i,j;
scanf("%d",&t);
dp[1][0][0]=dp[1][0][1]=1;
for(i=2;i<105;i++)
{
dp[i][0][0]=dp[i-1][0][0]+dp[i-1][0][1];
dp[i][0][1]=dp[i-1][0][0];
for(j=1;j<i;j++)
{
dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1];
dp[i][j][1]=dp[i-1][j-1][1]+dp[i-1][j][0];
}
}
while(t--)
{
int n,p;
scanf("%d%d",&n,&p);
printf("%d\n",dp[n][p][0]+dp[n][p][1]);
}
}