Codeforces Round #524 Div. 2 翻车记

　　A：签到。room里有一个用for写的，hack了一发1e8 1，结果用了大概600+ms跑过去了。惨绝人寰。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
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;
}
int n,m;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read(),m=read();
    cout<<(n*-)/m++(n*-)/m++(n*-)/m+;
    return ;
}

　　B：讨论一发即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
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;
}
int n;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    while (n--)
    {
        int x=read(),y=read();
        int p=y;if (y-x+&) y--;
        int ans=x&?(y-x+>>):-(y-x+>>);
        if (p-x+&) ans+=p&?-p:p;
        printf("%d\n",ans);
    }
    return ;
}

　　C：对矩形求个交，冷静一下瞎算算就行了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
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;
}
int T,n,m;
ll area(int x1,int y1,int x2,int y2){if (x1>x2||y1>y2) return ;return 1ll*(x2-x1+)*(y2-y1+);}
ll calcwhite(int x1,int y1,int x2,int y2)
{
    ll s=area(x1,y1,x2,y2);
    if (x1+y1&) return s>>;
    else return s+>>;
}
ll calcblack(int x1,int y1,int x2,int y2){return area(x1,y1,x2,y2)-calcwhite(x1,y1,x2,y2);}
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    T=read();
    while (T--)
    {
        n=read(),m=read();
        int x1=read(),y1=read(),x2=read(),y2=read();
        int x3=read(),y3=read(),x4=read(),y4=read();
        int x5=max(x1,x3),y5=max(y1,y3),x6=min(x2,x4),y6=min(y2,y4);
        ll white=calcwhite(,,n,m),black=calcblack(,,n,m);
        ll c1=calcblack(x1,y1,x2,y2);white+=c1,black-=c1;
        black+=calcwhite(x3,y3,x4,y4),white-=calcwhite(x3,y3,x4,y4);
        if (area(x5,y5,x6,y6)) black+=calcblack(x5,y5,x6,y6),white-=calcblack(x5,y5,x6,y6);
        printf("%I64d %I64d\n",white,black);
    }
    return ;
}

　　D：考虑答案为i时的最小和最大分裂次数是多少，各种递推式求通项到最后能捣鼓出来两个式子。显然答案若存在与n相差不会很大，暴力枚举check即可。当然式子不能直接算，需要各种乱搞防溢出。数学太差推一年。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
ll read()
{
    ll 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;
}
int T,n,m;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    T=read();
    while (T--)
    {
        ll n=read(),m=read();
        int ans=-;
        for (int i=(n>=?n-:);i<n;i++)
        {
            ll k=;
            for (int j=;j<=n-i+;j++) k<<=;
            k-=n-i;k-=;
            if (k<=m)
            {
                if (n+i+>=) {ans=i;break;}
                ll k=;for (int j=;j<=n-i-;j++) k<<=;k--;ll t=k+;
                for (int j=;j<=n+i+;j++)
                {
                    k<<=;
                    if (k>=*m+) {ans=i;break;}
                }
                if (ans!=-) break;
                t<<=;if (t>=*m+) {ans=i;break;}
                t<<=;if (t>=*m+) {ans=i;break;}
                k+=t;if (k>=*m+) {ans=i;break;}
                t=;
                for (int j=;j<=*i;j++)
                {
                    t<<=;
                    if (t>=*m+) {ans=i;break;}
                }
                if (ans!=-) break;
                k+=t;if (k>=*m+) {ans=i;break;}
            }
        }
        if (ans==-) printf("NO\n");
        else printf("YES %d\n",ans);
    }
    return ;
}

　　E：这个条件比较显然的等价于子矩形中每一行都能重排成回文串（即出现次数为奇数的字母不超过一个）且对称行的字母组成相同。预处理每一行的每一段是否合法以及字母出现次数的哈希值，暴力枚举子矩形左右端点，对上下端点的每段合法区间马拉车即可。完全没时间写了。

　　F：没看

　　小号打的。result：rank 139 rating +92