hdoj 3555 Bomb(DFA+dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

思路分析：该问题要求求解1—N中的数中含有49的数的个数，可以使用DFA来递推dp公式；详细解释点击链接查看；

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int MAX_N =  + ;
long long dp[MAX_N][];
int digit[MAX_N];

int NextState(int cur_state, int next_char)
{
    if (cur_state == )
    {
        if (next_char == )
            cur_state++;
    }
    else if (cur_state == )
    {
        if (next_char == )
            ++cur_state;
        else if (next_char != )
            --cur_state;
    }
    return cur_state;
}

int main()
{
    int case_times;
    long long n;

    scanf("%d", &case_times);
    while (case_times--)
    {
        int cur_state = , count = ;
        int len = ;
        long long ans = , temp_value = ;

        scanf("%I64d", &n);
        temp_value = ++n;
        while (temp_value)
        {
            digit[++len] = temp_value % ;
            temp_value /= ;
        }

        for (int i = len; i >= ; -- i)
        {
            ++count;
            for (int j = ; j < digit[i]; ++ j)
            {
                memset(dp, , sizeof(dp));
                dp[count][NextState(cur_state, j)] = ;
                for (int k = count + ; k <= len; ++ k)
                {
                    dp[k][] =  * dp[k-][] +  * dp[k-][];
                    dp[k][] = dp[k-][] + dp[k-][];
                    dp[k][] = dp[k-][] +  * dp[k-][];
                }
                ans += dp[len][];
            }
            cur_state = NextState(cur_state, digit[i]);
        }
        printf("%I64d\n", ans);
    }
    return ;
}