Codeforces 118D Caesar's Legions

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 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
解题思路:dp[i][j][0] 表示前i个人中有j个男生且最后一个人是男生的方案数,  dp[i][j][1] 表示前i个人有j个男生且最后一个人是女生的方案数，则我们最终求得的结果为dp[n1+n2][n1][0]+dp[n1+n2][n1][1]，状态转移见代码。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
const int mod = 100000000;
int dp[210][110][2];

int main() {

//freopen("aa.in", "r", stdin);

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