The Debut Album URAL - 2018[滚动数组递推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 10 9+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

Tags:  dynamic programming   ( hide tags for unsolved problems)

Difficulty: 102     Printable version     Submit solution     Discussion (8)
All submissions (5782)     All accepted submissions (1930)     Solutions rating (1293)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll mod = 1e9+7;
ll a, b, dp[2][2][307];
int n;
/*
    问题：求满足要求的混合音符的种类数
    当前状态f[i][j][k] : 表示 第i个为j{0, 1}时，连续k个后缀是j;
    那么
    k : 1 -> a
    dp[i][0][k] += dp[i - 1][0][k - 1];
    dp[i][1][1] += dp[i - 1][0][k];
    k : 1 -> b
    dp[i][1][k] += dp[i - 1][1][k - 1];
    dp[i][0][1] += dp[i - 1][1][k];
    这样50000*2*300太大了，所以用滚动数组。。。
    上面的状态转移方程我们可以用滚动数组2*2*300;
*/



ll solve()
{
    memset(dp, 0, sizeof(dp));
    dp[1][1][1] = 1;
    dp[1][0][1] = 1;
    for(int i = 2; i <= n; i++)
    {
        int now = i&1, pre = (i - 1)&1;
        memset(dp[now], 0, sizeof(dp[now])); //将当前行清空。。。
        for(int j = 1; j <= a; j++)
        {
            dp[now][0][j] = (dp[now][0][j] + dp[pre][0][j-1])%mod;
            dp[now][1][1] = (dp[now][1][1] + dp[pre][0][j])%mod;
        }
        for(int j = 1; j <= b; j++)
        {
            dp[now][1][j] = (dp[now][1][j] + dp[pre][1][j-1])%mod;
            dp[now][0][1] = (dp[now][0][1] + dp[pre][1][j])%mod;
        }
    }
    ll ans = 0;
    int now = n&1;
    for(int i = 1; i <= a; i++) ans = (ans + dp[now][0][i])%mod;
    for(int i = 1; i <= b; i++) ans = (ans + dp[now][1][i])%mod;
    return ans;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d%lld%lld", &n, &a, &b))
    {
        printf("%lld\n", solve());
    }
    return 0;
}