1487: Adjacent Bit Counts
Time Limit: 1 Sec
Memory Limit: 128 MB
Submit: 50
Solved: 20
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：
给你一个01串，规定01串的值为x1*x2 + x2*x3 + x3*x 4 + … + xn-1*x n，
现在给你两个数n,p，n表示01串的长度，p为01串的目标值。现在问你有几种组合方法使得长度为n的01串计算的值等于p。
例如 5 2
有6种组合方法：11100, 01110, 00111, 10111, 11101, 11011
//思路：一个标准的01串DP。
dp[n][p][0]+dp[n][p]+[1]即为长度为n，目标值为p的01串的组合方法数。
其中n表示串长，p表示目标值，0表示最后一位是0,1表示最后一位是1.
所以状态转移方程为：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];
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[110][110][2];
int main()
{
int t,i,j,k;
int l,n;
dp[1][0][0]=1;
dp[1][0][1]=1;
for(i=2;i<=100;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][0]+dp[i-1][j-1][1];
}
}
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&l,&n);
printf("%d\n",dp[l][n][0]+dp[l][n][1]);
}
return 0;
}
Source
