URAL - 2018 The Debut Album dp

2018. The Debut Album

Time limit: 2.0 second
Memory limit: 64 MB

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	output
3 2 1	4

Notes

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

Problem Author: Olga Soboleva (prepared by Alex Samsonov)
Problem Source: NEERC 2014, Eastern subregional contest

题意是给你n,a,b,构造一个由1和2构成的长度为n的序列，其中不能有连续a个1，和连续b个2

问有多少种这样的序列

刚开始题意看错了，一直想推公式，最后发现。。。。。。。。

dp【i】【j】表示长度为i并且第i个字符为j的序列个数                

显然j只有1和2两个取值

dp【i】【1】=sum（dp【i-k】【2】）（k<=a）;

dp【i】【2】=sum（dp【i-k】【1】）（k<=b）                   

因为这个位置是2，前面只能是前一个位置是1或前一个位置是2并且前两个位置是1或者。。。。。前b-1个位置是2并且前b个位置是1

#include<bits/stdc++.h>
using namespace std;
long long dp[500005][2];
const long long mo=1e9+7;
int main()
{
    int n,a,b,i,j;
    cin>>n>>a>>b;
    dp[1][1]=dp[1][2]=1;
    dp[0][1]=dp[0][2]=1;
    for(i=2;i<=n;i++)
    {
        for(j=1;j<=a;j++)
        {
            if(i-j<0)
                break;
            dp[i][1]=(dp[i][1]+dp[i-j][2])%mo;
        }
        for(j=1;j<=b;j++)
        {
            if(i-j<0)
                break;
            dp[i][2]=(dp[i-j][1]+dp[i][2])%mo;
        }
    }
  //  cout<<dp[n][1]<<" "<<dp[n][2]<<endl;
    cout<<(dp[n][1]+dp[n][2])%mo<<endl;
    return 0;
}