Adjacent Bit Counts
Time Limit: 1 Sec Memory Limit: 128 MB
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.
题目链接 http://acm.sdibt.edu.cn/vjudge/contest/view.action?cid=2099#problem/F
题意： 给你一个长度为 n 的 0/1 串，然后根据公式求答案为 k 的串的个数。
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]。
把长度为 n 的串 转化为长度为 n-1 的末尾为 0 或 1 的两种子问题。
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define eps 1e-9
#define MAXN 1e5+10
const int N=105;
int dp[110][110][2];
int main()
{
int i,j,k;
dp[1][0][1]=1;
dp[1][0][0]=1;
for(i=2;i<=N;i++)
{
dp[i][0][1]=dp[i-1][0][0];
dp[i][0][0]=dp[i-1][0][0]+dp[i-1][0][1];
for(j=1;j<=N;j++)
{
dp[i][j][1]=dp[i-1][j][0] + dp[i-1][j-1][1];
dp[i][j][0]=dp[i-1][j][0] + dp[i-1][j][1];
}
}
int T;
cin>>T;
while(T--)
{
int cnt,a,b;
cin>>cnt>>a>>b;
cout<<cnt<<" "<<dp[a][b][0]+dp[a][b][1]<<endl;
}
return 0;
}