hdu2089

基本的数位dp

#include <cstdio>

#include <cstring>

using namespace std;



#define D(x) x



const int MAX_DIGIT = 35;



int n, m;

int f[MAX_DIGIT];

int memoize[MAX_DIGIT][2][2][2];



void to_digits(int a)

{

    for (int i = 0; i < MAX_DIGIT; i++)

    {

        f[i] = a % 10;

        a /= 10;

    }

}



int dfs(int digit, bool less, bool contain, bool six)

{

    if (digit == -1)

    {

        return !contain;

    }

    if (memoize[digit][less][contain][six] != -1)

    {

        return memoize[digit][less][contain][six];

    }

    int limit = less ? 9 : f[digit];

    int ret = 0;

    for (int i = 0; i <= limit; i++)

    {

        if (i == 4 || (six && i == 2))

        {

            ret += dfs(digit - 1, less || i < f[digit], true, false);

            continue;

        }

        if (i == 6)

        {

            ret += dfs(digit - 1, less || i < f[digit], contain, true);

            continue;

        }

        ret += dfs(digit - 1, less || i < f[digit], contain, false);

    }

    memoize[digit][less][contain][six] = ret;

    return ret;

}



int main()

{

    while (scanf("%d%d", &n, &m), n | m)

    {

        n--;

        to_digits(n);

        memset(memoize, -1, sizeof(memoize));

        int ans_n = dfs(32, false, false, false);

        to_digits(m);

        memset(memoize, -1, sizeof(memoize));

        int ans_m = dfs(32, false, false, false);

        printf("%d\n", ans_m - ans_n);

    }

    return 0;

}

View Code