URAL 2018 The Debut Album DP中滚动数组用法

题目描述：

Description

Pop-group “Pink elephant” entered on recording their debut album. In fact they have only two songs: “My love” and “I miss you”, but each of them has a large number of remixes.

The producer of the group said that the album should consist of n remixes. On second thoughts the musicians decided that the album will be of interest only if there are no more than a remixes on “My love” in a row and no more than b remixes on “I miss you” in a row. Otherwise, there is a risk that even the most devoted fans won’t listen to the disk up to the end.

How many different variants to record the album of interest from n remixes exist? A variant is a sequence of integers 1 and 2, where ones denote remixes on “My love” and twos denote remixes on “I miss you”. Two variants are considered different if for some i in one variant at i-th place stands one and in another variant at the same place stands two.

Input

The only line contains integers n, a, b (1 ≤ a, b ≤ 300; max( a, b) + 1 ≤ n ≤ 50 000).

Output

Output the number of different record variants modulo 10 9+7.

Sample Input

input

3 2 1

output

4

Notes

In the example there are the following record variants: 112, 121, 211, 212.

题目分析：

n个位置，有两种物品，其中第一个物品不能连续超过a个，第二个物品不能连续超过b个。

很明显的DP题。任何人都能想到dp[i][j][k]来存状态，其中在i位置，选择j种物品，其中有k个连续的该物品。

但是，我们看n的范围是[1,50000]，a和b的范围是[1,300]。所以这个dp数组是50000*2*300的范围，肯定是MLE的。

这个时候就需要用到滚动数组了。

因为我们可以知道，这个dp状态转移方程中，下一状态只与其上一个状态有关，那么我们其实可以将i位置范围改成2，变成一个dp[2][2][300]的dp数组，这样就不会爆内存了。而关于这个滚动数组的用法，具体思想可以参考http://blog.csdn.net/niushuai666/article/details/6677982。

如果不看滚动数组，dp的转移方程也很容易推导的。我就不多赘述了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int MOD=1e9+7;

int n,a,b;
LL dp[2][2][310];
int ans;
int main()
{
while(scanf("%d%d%d",&n,&a,&b)!=-1)
{
memset(dp,0,sizeof(dp));
dp[0][0][0]=1;
dp[0][1][0]=1;
ans=0;
for(int i=1; i<=n; i++)
{
int now=i&1;
int pre=now^1;
//            if (now==1 && pre==0)
//            printf("%d %d\n",pre+i-1,now+i-1);
//            else if (now==0 && pre==1)
//            printf("%d %d\n",pre+i-2,now+i);//原数组位置
memset(dp[now],0,sizeof(dp[now]));
for(int j=1; j<=a; j++)
{
dp[now][0][j]+=dp[pre][0][j-1];
dp[now][0][j]%=MOD;
dp[now][1][1]+=dp[pre][0][j];
dp[now][1][1]%=MOD;
}
for(int j=1; j<=b; j++)
{
dp[now][1][j]+=dp[pre][1][j-1];
dp[now][1][j]%=MOD;
dp[now][0][1]+=dp[pre][1][j];
dp[now][0][1]%=MOD;
}
}
for(int i=1; i<=a; i++)
{
ans+=dp[n&1][0][i];
//printf("%I64d\n",dp[n&1][0][i]);
ans%=MOD;
}
for(int i=1; i<=b; i++)
{
ans+=dp[n&1][1][i];
//printf("%I64d\n",dp[n&1][1][i]);
ans%=MOD;
}
printf("%d\n",ans);
}
return 0;
}