ural 2018. The Debut Album 滚动数组dp

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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

inputoutput

3 2 1

4

Hint

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

构造一个仅有1和2组成的长度为n的串，要求连续1的个数不能超过a，连续2的个数不能超过b，求一共有多少种符合情况的串。

参考：点击打开链接

//0.562210 KB#include<stdio.h>#include<string.h>#include<algorithm>#define M 1000000007#define ll long longusing namespace std;ll dp[2][2][307];int main(){    int n,a,b,now,pre;    while(scanf("%d%d%d",&n,&a,&b)!=EOF)    {        memset(dp,0,sizeof(dp));        dp[0][0][0]=1;dp[0][1][0]=1;        for(int i=1;i<=n;i++)        {            now=i&1,pre=(i-1)&1;            memset(dp[now],0,sizeof(dp[now]));            for(int j=1;j<=a;j++)            {                dp[now][0][j]+=dp[pre][0][j-1];                if(dp[now][0][j]>M)dp[now][0][j]%=M;                dp[now][1][1]+=dp[pre][0][j];                if(dp[now][1][1]>M)dp[now][1][1]%=M;            }            for(int j=1;j<=b;j++)            {                dp[now][1][j]+=dp[pre][1][j-1];                if(dp[now][1][j]>M)dp[now][1][j]%=M;                dp[now][0][1]+=dp[pre][1][j];                if(dp[now][0][1]>M)dp[now][0][1]%=M;            }        }        ll ans=0;        for(int i=1;i<=a;i++)        {            ans+=dp[n&1][0][i];            if(ans>M)ans%=M;        }        for(int i=1;i<=b;i++)        {            ans+=dp[n&1][1][i];            if(ans>M)ans%=M;        }        printf("%lld\n",ans);    }    return 0;}