Codeforces Round #524 Div. 2 翻车记

Posted 2021-01-22 gloid

　　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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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*2-1)/m+1+(n*5-1)/m+1+(n*8-1)/m+1;
    return 0;
}

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　　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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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+1&1) y--;
        int ans=x&1?(y-x+1>>1):-(y-x+1>>1);
        if (p-x+1&1) ans+=p&1?-p:p;
        printf("%d
",ans);
    }
    return 0;
}

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　　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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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 0;return 1ll*(x2-x1+1)*(y2-y1+1);}
ll calcwhite(int x1,int y1,int x2,int y2)
{
    ll s=area(x1,y1,x2,y2);
    if (x1+y1&1) return s>>1;
    else return s+1>>1;
}
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(1,1,n,m),black=calcblack(1,1,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
",white,black);
    }
    return 0;
}

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　　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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
    ll x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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=-1;
        for (int i=(n>=63?n-63:0);i<n;i++)
        {
            ll k=1;
            for (int j=1;j<=n-i+1;j++) k<<=1;
            k-=n-i;k-=2;
            if (k<=m)
            {
                if (n+i+1>=63) {ans=i;break;}
                ll k=1;for (int j=1;j<=n-i-1;j++) k<<=1;k--;ll t=k+1;
                for (int j=1;j<=n+i+1;j++)
                {
                    k<<=1;
                    if (k>=3*m+2) {ans=i;break;}
                }
                if (ans!=-1) break;
                t<<=1;if (t>=3*m+2) {ans=i;break;}
                t<<=1;if (t>=3*m+2) {ans=i;break;}
                k+=t;if (k>=3*m+2) {ans=i;break;}
                t=1;
                for (int j=1;j<=2*i;j++)
                {
                    t<<=1;
                    if (t>=3*m+2) {ans=i;break;}
                }
                if (ans!=-1) break;
                k+=t;if (k>=3*m+2) {ans=i;break;}
            }
        }
        if (ans==-1) printf("NO
");
        else printf("YES %d
",ans);
    }
    return 0;
}

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　　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