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

input	output
3 2 1	4

Notes

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

Hint

题意：

在n个位置填充1 or 2. 要求连续的1的个数不超过a，连续的2的个数不超过b. 求方案数.

思路：

用一个二维数组，来进行存储方案数，其实只需要第i个位置后连续的j（<a||<b）个位置填1或2,那么我们可以dp[i+j][1] 表示第i个位置填1，在i位置后面连续填j个2的方案个数 dp[i+j][2] 表示第i个位置填2，在i位置后面连续填j个1的方案个数 !!!!!

代码：

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+2;
const int mod=1e9+7;
int main()
{
long long dp
[3];
memset(dp,0,sizeof(dp));
int i,j,a,b,n;
cin>>n>>a>>b;
dp[0][1]=dp[0][2]=1;
for(i=0;i<=n;i++)
{
for(j=1;j<=a&&i+j<=n;j++)
dp[i+j][1]=(dp[i+j][1]+dp[i][2])%mod;
for(j=1;j<=b&&i+j<=n;j++)
dp[i+j][2]=(dp[i+j][2]+dp[i][1])%mod;

}
cout<<(dp
[1]+dp
[2])%mod<<endl;
return 0;
}
提问：：：

求哪个大神告诉我，为什么对它进行求余1e9+7就能AC？？？？好困惑，菜鸟一个