Codeforce 题目118D. Caesar's Legions（DP背包）

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 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
题目大意：n个1，m个2，1最多k1个连续，2最多k2个连续，问有多少种情况

ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define mod 100000000using namespace std;
int dp[220][220][3];
int n,m,k1,k2;
int DP()
{
memset(dp,0,sizeof(dp));
int i,j,t;
dp[0][0][0]=1;
dp[0][0][1]=1;
for(i=1;i<=n+m;i++)
{
for(j=min(i,n);j>=1;j--)
{
for(t=1;t<=j&&t<=k1;t++)
{
if(i-j<=m)
dp[i][j][0]=(dp[i][j][0]+dp[i-t][i-j][1])%mod;
}
}
for(j=min(i,m);j>=1;j--)
{
for(t=1;t<=j&&t<=k2;t++)
{
if(i-j<=n)
dp[i][j][1]=(dp[i][j][1]+dp[i-t][i-j][0])%mod;
}
}
}
return (dp[n+m]
[0]+dp[n+m][m][1])%mod;
}
int main()
{
//int n,m,k1,k2;
while(scanf("%d%d%d%d",&n,&m,&k1,&k2)!=EOF)
{
printf("%d\n",DP());
}
}