ural 2018. The Debut Album 滚动数组dp

最新推荐文章于 2019-08-07 20:38:50 发布

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最新推荐文章于 2019-08-07 20:38:50 发布

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解决一个关于专辑录制的问题，其中包含两种歌曲的不同混音版本，需要确保连续相同歌曲的混音数量不超过一定限制。

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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 10 9+7.

Sample

input	output
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。

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//0.562	210 KB
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 1000000007
#define ll long long
using 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;
}