河南第六届ACM省赛（Adjacent Bit Counts）

题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=715

思路：dp[i][j][k]，i为位数，j为和,k为最后一位是啥

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>

const int inf = 0x7f7f7f7f;//2139062143
typedef long long ll;
using namespace std;

ll dp[110][110][2];

int main()
{
    int t;
    dp[1][0][1] = dp[1][0][0] = 1;
    for(int 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(int j=1; j<i; j++)
        {
            dp[i][j][0] = dp[i-1][j][1] + dp[i-1][j][0];
            dp[i][j][1] = dp[i-1][j][0] + dp[i-1][j-1][1];
        }
    }
    scanf("%d",&t);
    while(t--)
    {
        int n,p;
        scanf("%d%d",&n,&p);
        ll sum = dp[n][p][0] + dp[n][p][1];
        printf("%I64d\n",sum);
    }
    return 0;
}