URAL 2018. The Debut Album （dp）__動態規劃

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

題意

一個長度為 n 的數列，其中 1 連續的個數不能超過 a ， 2 連續的個數不能超過 b ，問總共有多少個這樣的數列。

思路

dp[i][k] 代表長度為 i 的數列，以 k 結尾並滿足題意的個數。

則有： dp[i][k]+=dp[i-j][k^1]

其中 dp[i-j][k^1] 代表當前點 i 之前最遠距離為 a||b 的那一點與當前點不同的情況，因為在這個範圍以內都滿足題意。

AC 代碼

#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;}