URAL2018——The Debut Album(DP)

                             G - The Debut Album

 

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.

Example

inputoutput

3 2 1

4

Notes

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

给你一个串的长度n,串由1,2组成，1连续不超过a个，2连续不超过b个，问有多少种这样的串,对1e9+7取模。

一看题就觉得记忆化可搞，几分钟就写出来了，奈何没考虑空间复杂度，ME。。

换了个思路，干脆两个DP，dp1[i]表示第i个为1的串的方案数，dp2[j]表示第j个为2的串的方案数。

状态转移方程dp1[i]+=dp2[i-j] dp2[i]+=dp1[i-j]  最后答案dp1[n]+dp2[n]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
using namespace std;
typedef long long ll;
#define PI 3.1415926535897932
#define E 2.718281828459045
#define INF 0xffffffff//0x3f3f3f3f
#define mod 1000000007
const int MOD=1e9+7;

const int M=1005;
int n,m;
int a,b;
int dp1[50001],dp2[50001];
int main()
{
    int i,j,k,t;
    int cas=0;
    int ans=0;
    //scanf("%d",&t);
    while(~scanf("%d%d%d",&n,&a,&b))
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        dp1[0]=dp2[0]=1;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=a&&j<=i; j++)
                dp1[i]=(dp1[i]+dp2[i-j])%MOD;
            for(j=1; j<=b&&j<=i; j++)
                dp2[i]=(dp2[i]+dp1[i-j])%MOD;
        }
        printf("%d\n",(dp1[n]+dp2[n])%MOD);
    }
    return 0;
}