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 = 20 + 10;
long long dp[MAX_N][3];
int digit[MAX_N];

int NextState(int cur_state, int next_char)
{
    if (cur_state == 0)
    {
        if (next_char == 4)
            cur_state++;
    }
    else if (cur_state == 1)
    {
        if (next_char == 9)
            ++cur_state;
        else if (next_char != 4)
            --cur_state;
    }
    return cur_state;
}

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

    scanf("%d", &case_times);
    while (case_times--)
    {
        int cur_state = 0, count = 0;
        int len = 0;
        long long ans = 0, temp_value = 0;

        scanf("%I64d", &n);
        temp_value = ++n;
        while (temp_value)
        {
            digit[++len] = temp_value % 10;
            temp_value /= 10;
        }

        for (int i = len; i >= 1; -- i)
        {
            ++count;
            for (int j = 0; j < digit[i]; ++ j)
            {
                memset(dp, 0, sizeof(dp));
                dp[count][NextState(cur_state, j)] = 1;
                for (int k = count + 1; k <= len; ++ k)
                {
                    dp[k][0] = 9 * dp[k-1][0] + 8 * dp[k-1][1];
                    dp[k][1] = dp[k-1][0] + dp[k-1][1];
                    dp[k][2] = dp[k-1][1] + 10 * dp[k-1][2];
                }
                ans += dp[len][2];
            }
            cur_state = NextState(cur_state, digit[i]);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}