URAL 2018. The Debut Album （dp）

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 109+7.

Sample input

3 2 1

Sample output

4

题意

一个长度为

n

的数列，其中

1

连续的个数不能超过

a

，

2

连续的个数不能超过

b

，问总共有多少个这样的数列。

思路

dp[i][k]

代表长度为

i

的数列，以

k

结尾并满足题意的个数。

则有：

dp[i][k]+=dp[i-j][k^1]

其中

dp[i-j][k^1]

代表当前点

i

之前最远距离为

a||b

的那一点与当前点不同的情况，因为在这个范围以内都满足题意。

AC 代码

#include<bits/stdc++.h>
using namespace std;

const int mod=1e9+7;

int dp[51000][2];

int main()
{
int n,a,b;
cin>>n>>a>>b;
dp[0][0]=dp[0][1]=1;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=min(i,a); j++)
dp[i][0]=(dp[i][0]+dp[i-j][1])%mod;
for(int j=1; j<=min(i,b); j++)
dp[i][1]=(dp[i][1]+dp[i-j][0])%mod;
}
cout<<(dp
[0]+dp
[1])%mod;
return 0;
}