<背包>solution-CF118D_Caesar's Legions

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.

Examples

input1

2 1 1 10

output1

1

input2

2 3 1 2

output2

5

input3

2 4 1 1

output3

0

人话：有n1个步兵和n2个骑兵要排成一排,连续的步兵数量不能超过k1个,连续的骑兵数量不能超过k2个,问有几种排列方案

首先，这肯定是个dp

我一开始考虑用\(f[i][j]\)表示i个步兵，j个骑兵的方案数，然后发现列不出方程。。。

这时候，我们可以考虑加一维来使我们能列方程

\(f[i][j][0/1]\)表示i个步兵，j个骑兵，当前这一个是0：步兵；1：骑兵的方案数

这时候方程就很好列了，根据题意

\(f[i][j][0]\)只可能从\(f[i-k][j][1]\)转移，这时候\(0<k<k1\) ，

\(f[i][j][1]\)只可能从\(f[i][j-k][0]\)转移，这时候\(0<k<k2\)，

这里的\(k\)表示的是连续的k个步兵或连续的k个骑兵

然后，Code：

#include <cstdio>
#include <cctype>
#include <algorithm>
#define reg register
using namespace std;
const int MaxN=101;
const int p=100000000;
template <class t> inline void rd(t &s)
{
	s=0;
	reg char c=getchar();
	while(!isdigit(c))
		c=getchar();
	while(isdigit(c))
		s=(s<<3)+(s<<1)+(c^48),c=getchar();
	return;
}
int f[MaxN][MaxN][2];
signed main(void)
{
	int n1,n2,k1,k2;
	rd(n1);rd(n2);rd(k1);rd(k2);
	for(int i=1;i<=k1;++i)
		f[i][0][0]=1;
	for(int i=1;i<=k2;++i)
		f[0][i][1]=1;
	for(int i=1;i<=n1;++i)
		for(int j=1;j<=n2;++j)
		{
			for(int k=1;k<=min(i,k1);++k)
				f[i][j][0]+=f[i-k][j][1],f[i][j][0]%=p;
			for(int k=1;k<=min(j,k2);++k)
				f[i][j][1]+=f[i][j-k][0],f[i][j][1]%=p;
//			printf("%d %d\n",min(i,k1),min(j,k2));
//			printf("%d %d\n",f[i][j][0],f[i][j][1]);
		}
	printf("%d",(f[n1][n2][0]+f[n1][n2][1])%p);
	return 0;
}

有一点要注意的，for枚举k时要判一下，不能出现访问负数下标的情况，否则会RE或WA