Caesar's Legions(三维dp)

Caesar's Legions

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Gaius Julius Caesar, a famous general, loved to line up his soldiers. Overall the army had n₁ footmen and n₂ horsemen. Caesar thought that an arrangement is not beautiful if somewhere in the line there are strictly more that k₁ footmen standing successively one after another, or there are strictly more than k₂ horsemen standing successively one after another. Find the number of beautiful arrangements of the soldiers.

Note that all n₁ + n₂ warriors should be present at each arrangement. All footmen are considered indistinguishable among themselves. Similarly, all horsemen are considered indistinguishable among themselves.

Input

The only line contains four space-separated integers n₁, n₂, k₁, k₂ (1 ≤ n₁, n₂ ≤ 100, 1 ≤ k₁, k₂ ≤ 10) which represent how many footmen and horsemen there are and the largest acceptable number of footmen and horsemen standing in succession, correspondingly.

Output

Print the number of beautiful arrangements of the army modulo 100000000(10⁸). That is, print the number of such ways to line up the soldiers, that no more than k₁ footmen stand successively, and no more than k₂ horsemen stand successively.

Sample Input

Input

2 1 1 10

Output

1

Input

2 3 1 2

Output

5

Input

2 4 1 1

Output

0

Hint

Let's mark a footman as 1, and a horseman as 2.

In the first sample the only beautiful line-up is: 121

In the second sample 5 beautiful line-ups exist: 12122, 12212, 21212, 21221, 22121

 

The problem is solved lazy dynamics. Let z[n1] [n2] [2] - a number of ways to place troops in a legion of Caesar. Indicate the following parameters, n1 – is a number of footmen, n2 – is a number of horseman, the third parameter indicates what troops put Caesar in the beginning of the line. If Caesar wants to put the footmen, the state dynamics of the z [n1] [n2] [0] go to the state

z [n1 - i] [n2] [0], where 0 <= i <= min (k1, n1) . If Caesar wants to put the riders, the state dynamics of the z [n1] [n2] [1] go to the state z [n1] [n2 - i] [1], where 0 <= I <= min (k2, n2) .

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int mod = 100000000 ;
int n1 , n2 , k1 , k2 ;
int a[110][110][2] ;

int Caesar (int n1 , int n2 , int f)
{
    if (a[n1][n2][f] != -1 ) {
        return a[n1][n2][f] ;
    }
    if (n1 + n2 == 0) {
        a[n1][n2][f] = 1 % mod ;
        return a[n1][n2][f] ;
    }
    a[n1][n2][f] = 0 ;
    int i ;
    if (f == 0) {
        for (i = 1 ; i <= min (k1 , n1 ) ; i++) {
            a[n1][n2][f] += Caesar (n1 - i , n2 , 1 - f) ;
            a[n1][n2][f] %= mod ;
        }
    }
    else {
        for (i = 1 ; i <= min (k2 , n2 ) ; i++) {
            a[n1][n2][f] += Caesar (n1 , n2 - i , 1 - f ) ;
            a[n1][n2][f] %= mod ;
        }
    }
    return a[n1][n2][f] ;
}

void solve ()
{
    memset (a , 0xFF , sizeof(a) ) ;
    printf ("%d
" , ( Caesar (n1 , n2 , 0 ) + Caesar (n1 , n2 , 1 ) ) % mod ) ;
}

int main ()
{
#ifdef online_jude
    freopen ("a.txt" , "r" , stdin ) ;
#endif // online_jude
    scanf ("%d%d%d%d" , &n1 , &n2 , &k1 , &k2 ) ;
    solve () ;
    return 0 ;
}