Codeforces Beta Round #89 (Div. 2)D. Caesar's Legions

D. Caesar's Legions

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gaius Julius Caesar, a famous general, loved to line up his soldiers. Overall the army had n1 footmen and n2horsemen. Caesar thought that an arrangement is not beautiful if somewhere in the line there are strictly more thatk1 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 k2horsemen stand successively.

Sample test(s)

input

2 1 1 10

output

1

input

2 3 1 2

output

5

input

2 4 1 1

output

0

Note

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

題意：有n1個步兵，n2個騎兵，將這些士兵進行排隊，步兵不能連續排超過k1個，騎兵不能連續排超過k2個，問你排列

      的方案數。

思路：dp[i][j][0]表示排列i個步兵，j個騎兵滿足步兵要求的方案數：dp[i][j][0]+=dp[i-k][j][1](k表示連續放k個步兵          dp[i][j][1],表示滿足騎兵的方法)

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mod 100000000

int dp[105][105][2];
int main()
{
    int n1,n2,k1,k2;
    int i,j,k;
    while(scanf("%d%d%d%d",&n1,&n2,&k1,&k2)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0][0][0]=dp[0][0][1]=1;
        for(i=0;i<=n1;i++)
        {
            for(j=0;j<=n2;j++)
            {
                for(k=1;k<=k1&&k<=i;k++)
                {
                    dp[i][j][0]=(dp[i][j][0]+dp[i-k][j][1])%mod;
                }
                for(k=1;k<=k2&&k<=j;k++)
                {
                    dp[i][j][1]=(dp[i][j][1]+dp[i][j-k][0])%mod;
                }
            }
        }
        printf("%d\n",(dp[n1][n2][0]+dp[n1][n2][1])%mod);
    }
    return 0;
}