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 a2 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[n][0]+dp[n][1])%mod;
    return 0;
}