====== 2020/05/16 牛客假日团队赛41 ======
===== 比赛信息 =====
日期： 2020/05/16
链接： [[https://ac.nowcoder.com/acm/contest/5208#question]]\\
做题统计：王瑞琦： 冯宇扬： 常程：2\\
===== 题解 =====
==== A ====
solved by, upsolved by .\\

题意：\\
题解：
==== B photo ====
solved by 常程, upsolved by .\\

题意：N（$2\le N\le10^9$）头牛站成一排照相。有K（$1\le K\le1000$）对牛不想被照在同一张相片里。求最少照的次数。\\
题解：这道题在中途分界并没有能减少照的次数————不冲突的分配给那个区间都无所谓，冲突的必须分区间，因此直接贪心，发现矛盾就换区间。\\
<hidden code>
<code cpp>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
int n,k,line[2005],t=0,f[2005],ans=1;
map<int,int>to,vist;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=k;i++)
    {
        scanf("%d%d",&line[++t],&line[++t]);
        to[line[t-1]]=line[t];
        to[line[t]]=line[t-1];
    }
    sort(line+1,line+t+1);
    for(int i=1,v=1;i<=t;i++)
    {
        if(vist[to[line[i]]]==v)
        {
            v++;
            ans++;
        }
        vist[line[i]]=v;
    }
    printf("%d",ans);
    return 0;
}
</code>
</hidden>
==== C ====
solved by, upsolved by .\\

题意：\\
题解：
==== D ====
solved by, upsolved by .\\

题意：\\
题解：
==== E ====
solved by, upsolved by .\\

题意：\\
题解：
==== F ====
solved by, upsolved by .\\

题意：\\
题解：
==== G ====
solved by, upsolved by .\\

题意：\\
题解：
==== H ====
solved by, upsolved by .\\

题意：\\
题解：
==== I ====
solved by 常程, upsolved by .\\

题意：N（$1\le N\le5\times10^4$）只牛排成一排，每只牛有一个id，id相同、距离不到K（$1\le K<N$）的两只牛称为"crowded"。求最大的"crowded"的牛的id。\\
题解：大水题。直接记录每一个id上一次出现的位置就好了。
<hidden code>
<code cpp>
#include<cstdio>
#include<map>
using namespace std;
int n,k,a,ans=-1;
map<int,int>last;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a);
        if(last[a]&&i-last[a]<=k&&a>ans)
            ans=a;
        last[a]=i;
    }
    printf("%d",ans);
    return 0;
}
</code>
</hidden>
==== J ====
solved by, upsolved by .\\

题意：\\
题解：
==== K ====
solved by, upsolved by .\\

题意：\\
题解：
==== L ====
solved by, upsolved by .\\

题意：\\
题解：

===== replay =====
===== 总结 =====