Light OJ 1032

数位dp，许多数位dp需要统计某种模式（子串）出现的数量，这种题通常需要在递归参数中加入高位已经出现过的模式的数量。

#include <cstdio>

#include <cstring>

using namespace std;



#define D(x) 



const int MAX_DIGIT = 40;



long long n;

int f[MAX_DIGIT];

long long memoize[MAX_DIGIT][2][MAX_DIGIT];

int cnt;



int to_digits(long long a)

{

    int ret = 0;

    while (a > 0)

    {

        f[ret++] = a % 2;

        a /= 2;

    }

    return ret;

}



long long dfs(int digit, bool less, bool last, int adj_num)

{

    D(cnt++);

    if (digit < 0)

    {

        return adj_num;

    }

    if (less && memoize[digit][last][adj_num] != -1)

    {

        return memoize[digit][last][adj_num];

    }

    int limit = less ? 1 : f[digit];

    long long ret = 0;

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

    {

        int delta = (i == 1 && last) ? 1 : 0;

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

    }

    if (less)

    {

        memoize[digit][last][adj_num] = ret;

    }

    return ret;

}



long long work(long long n)

{

    if (n < 0)

    {

        return 0;

    }

    if (n == 0)

    {

        return 0;

    }

    int len = to_digits(n);

    cnt = 0;

    return dfs(len - 1, false, false, 0);

}



int main()

{

    int t;

    scanf("%d", &t);

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

    for (int i = 1; i <= t; i++)

    {

        int a;

        scanf("%d", &a);

        printf("Case %d: %lld\n", i, work(a));

        D(printf("%d\n", cnt));

    }

    return 0;

}

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