uva 10157 expressions

Problem A: EXPRESSIONS 

Let X be the set of correctly built parenthesis expressions. The elements of X are strings consisting only of the characters �(� and �)�. The set X is defined as follows:

• an empty string belongs to X
• if A belongs to X, then (A) belongs to X
• if both A and B belong to X, then the concatenation AB belongs to X.

For example, the following strings are correctly built parenthesis expressions (and therefore belong to the set X):

()(())()

(()(()))

The expressions below are not correctly built parenthesis expressions (and are thus not in X):

(()))(()

())(()

Let E be a correctly built parenthesis expression (therefore E is a string belonging to X).

The length of E is the number of single parenthesis (characters) in E.

The depth D(E) of E is defined as follows:

               ì 0 if E is empty 
D(E)= í D(A)+1 if E = (A), and A is in X 
               î max(D(A),D(B)) if E = AB, and A, B are in X

For example, the length of �()(())()� is 8, and its depth is 2.

What is the number of correctly built parenthesis expressions of length n and depth d, for given positive integers n and d?

Task 
Write a program which

• reads two integers n and d
• computes the number of correctly built parenthesis expressions of length n and depth d;

Input data 
Input consists of lines of pairs of two integers - n and d, at most one pair on line, 2 £ n £ 300, 1£ d £ 150. 
The number of lines in the input file is at most 20, the input may contain empty lines, which you don't need to consider.

Output data 
For every pair of integers in the input write single integer on one line - the number of correctly built parenthesis expressions of length n and depth d.

Example 
Input data                                   Output data 
6 2                 3 
300 150             1 

There are exactly three correctly built parenthesis expressions of length 6 and depth 2: 
(())() 
()(()) 
(()()) 

#include <cstdio>#include <cstring>#include <cassert>const int LEN = 100;const int NMAX = 300;const int DMAX = 150;int f[NMAX + 1][DMAX + 1][LEN];bool isCalculated[NMAX + 1][DMAX + 1];void add (int a[], int b[]){    int carry = 0;    for (int i=0; i<LEN; ++i) {        int n = (carry + a[i] + b[i]) % 10;        carry = (carry + a[i] + b[i]) / 10;        a[i] = n;    }    assert (carry == 0);}void substract (int a[], int b[]){    int borrow = 0;    for (int i=0; i<LEN; ++i) {        int n = (a[i] - b[i] + borrow + 10) % 10;        borrow = (a[i] - b[i] + borrow + 10) / 10 - 1;        a[i] = n;    }    assert (borrow == 0);}void multiply (int a[], int b[]){    static int c[LEN];    memset (c, 0, sizeof (c));    int I = LEN;    while (--I >= 0 && a[I] == 0) {}    ++I;    int K = LEN;    while (--K >= 0 && b[K] == 0) {}    ++K;    for (int i=0; i<I; ++i) {        for (int k=0; k<K && i+k<LEN; ++k) {            c[i + k] += a[i] * b[k];        }    }    int carry = 0;    for (int i=0; i<LEN; ++i) {        int n = (carry + c[i]) % 10;        carry = (carry + c[i]) / 10;        c[i] = n;    }    assert (carry == 0);    memcpy (a, c, sizeof (c));}void print (int a[]){    int i = LEN;    while (--i > 0 && a[i] == 0) {}    while (i >= 0) {        printf ("%d", a[i--]);    }    printf ("\n");}void calculate (int n, int d){#ifndef ONLINE_JUDGE    printf ("calculate (%d, %d)\n", n, d);#endif    if (isCalculated[n][d]) {        return;    }    isCalculated[n][d] = true;    for (int i=0; i<LEN; ++i) {        f[n][d][i] = 0;    }    if (n % 2 == 1) {        return;    }    if (n == 0) {        f[n][d][0] = 1;        return;    }    if (d == 0) {        return;    }    static int t[LEN];    for (int i=0; i<n; i+=2) {        assert (n - 2 - i >= 0);        assert (d - 1 >= 0);        calculate (i, d);        calculate (n - 2 - i, d - 1);        memcpy (t, f[i][d], sizeof (t));        multiply (t, f[n - 2 - i][d - 1]);        add (f[n][d], t);    }#ifndef ONLINE_JUDGE    printf ("f[%d][%d] = ", n, d);    print (f[n][d]);#endif}int main (int argc, char *argv[]){    int n, d;    while (scanf ("%d%d", &n, &d) == 2) {        if (d == 0 && n == 0) {            printf ("1\n");            continue;        } else if (d == 0) {            printf ("0\n");            continue;        }        calculate (n, d);        calculate (n, d - 1);        static int ans[LEN];        memcpy (ans, f[n][d], sizeof (ans));        substract (ans, f[n][d - 1]);        print (ans);    }    return 0;}