Codeforces Round#509 Div.2翻车记

　　A：签到

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
int n,a[N];
int main()
{
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    sort(a+,a+n+);
    cout<<a[n]-a[]+-n;
    return ;
}

　　B：签到

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define ll long long
ll a,b,x,y;
ll gcd(ll n,ll m){return m==?n:gcd(m,n%m);}
int main()
{
    cin>>a>>b>>x>>y;
    ll tmp=gcd(x,y);x/=tmp,y/=tmp;
    cout<<min(a/x,b/y);
    return ;
}

　　C：贪心，每次选择最后一次喝咖啡最早的一天，看是否能将此次喝咖啡加入这天，不行就新开一天。堆维护即可。开始的时候写了半天线性的一直wa on 9，掉rating瞬间稳了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
int n,m,k,tot,ans[N];
struct data{int x,i;
    bool operator <(const data&a) const
    {
        return x<a.x;
    }
}a[N];
struct data2{int id,tim;
    bool operator <(const data2&a) const
    {
        return tim>a.tim;
    }
};
priority_queue<data2> q;
int main()
{
    n=read(),m=read(),k=read()+;
    for (int i=;i<=n;i++) a[i].x=read(),a[i].i=i;
    sort(a+,a+n+);
    for (int i=;i<=n;i++)
    {
        if (q.empty()||q.top().tim+k>a[i].x)
        {
            tot++;ans[a[i].i]=tot;
            q.push((data2){tot,a[i].x});
        }
        else
        {
            ans[a[i].i]=q.top().id;
            data2 y=q.top();q.pop();
            q.push((data2){y.id,a[i].x});
        }
    }
    cout<<tot<<endl;
    for (int i=;i<=n;i++) printf("%d ",ans[i]);
    return ;
}

　　D：按间隔长度尺取即可。没看到没有相交段，惨惨。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 200010
int n,m,cnt,c[N],d[N],ans,t=;
struct data{int x,y;
    bool operator <(const data&a) const
    {
        return x<a.x;
    }
}a[N],b[N];
int main()
{
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i].x=read(),a[i].y=read();
    sort(a+,a+n+);
    for (int i=;i<=n;i++)
    {
        int x=a[i].y,t=i;
        while (t<n&&a[t+].x<=x) t++,x=max(x,a[t].y);
        cnt++,b[cnt].x=a[i].x,b[cnt].y=x;
        i=t;
    }
    for (int i=;i<cnt;i++)
    t++,c[t]=b[i].y-b[i].x,d[t]=b[i+].x-b[i].y;
    c[++t]=b[cnt].y-b[cnt].x;
    for (int i=;i<=t;i++) c[i]+=c[i-];
    for (int i=;i<t;i++) d[i]+=d[i-];
    ans=c[]+m;
    int x=;
    for (int i=;i<t;i++)
    {
        while (x+<t&&d[x+]-d[i-]<m) x++;
        ans=max(ans,c[x+]-c[i-]+m);
    }
    cout<<ans;
}

　　E：容易发现任意一条边的两端总有一端包含最大值，先把这个判掉。不妨设n号点为树根，此时考虑n-1号点，其与n号点的距离就应为n-1号点的出现次数。类似地发现构造一个链或者菊花就好了，构造完再判断是否合法。n<=1000可能是为了写起来舒服一点。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1010
int n,a[N],b[N],cnt[N],tree[N],x;
int main()
{
    n=read();
    for (int i=;i<n;i++)
    {
        a[i]=read(),b[i]=read();cnt[a[i]]++;
        if (b[i]!=n) {printf("NO\n");return ;}
    }
    tree[]=n;x=;
    for (int i=n-;i;i--)
    if (cnt[i]) tree[x+=cnt[i]]=i;
    for (int i=n-;i>=;i--)
    if (!cnt[i])
        for (int j=;j<=n;j++)
        if (!tree[j]) {tree[j]=i;break;}
    for (int i=;i<n;i++)
    {
        int x=;
        for (int j=i+;j<=n;j++) x=max(x,tree[j]);
        cnt[x]--;
        if (cnt[x]<) {printf("NO\n");return ;}
    }
    printf("YES\n");
    for (int i=;i<n;i++) cout<<tree[i]<<' '<<tree[i+]<<endl;
}

　　F：比赛时看错题了以为是找最长的交替等差子序列，这拿头做啊。结束半个小时了才发现。考虑如何选取公差，若d=a*b且b是奇数，那么d=a一定会更优。所以只枚举2的幂次作为公差然后map存一下取模的结果就行了。注意ans初值设成2，因为可能存在横坐标相同的点。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 100010
int n,m,x,y,a[N],b[N],ans=;
map<int,int> f,g;
int main()
{
    n=read(),x=read();
    for (int i=;i<=n;i++) a[i]=read();
    m=read(),y=read();
    for (int i=;i<=m;i++) b[i]=read();
    for (int x=;x<=1E9;x<<=)
    {
        f.clear();g.clear();
        for (int i=;i<=n;i++) f[a[i]%(x<<)]++;
        for (int i=;i<=m;i++) g[b[i]%(x<<)]++;
        for (int i=;i<=n;i++)
        ans=max(ans,f[a[i]%(x<<)]+g[(a[i]+x)%(x<<)]);
        for (int i=;i<=m;i++)
        ans=max(ans,g[b[i]%(x<<)]+f[(b[i]+x)%(x<<)]);
    }
    cout<<ans;
}

　　总体是一场readforces&speedforces。于是就翻车了。当然自己还是弱爆。