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====== 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。\\


#include
#include
#include
using namespace std;
int n,k,line[2005],t=0,f[2005],ans=1;
mapto,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;
}

==== 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


#include
#include
using namespace std;
int n,k,a,ans=-1;
maplast;
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;
}

==== 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 =====