Codeforces Round #524 Div. 2 翻車記

　　A：簽到。room里有一個用for寫的，hack了一發1e8 1，結果用了大概600+ms跑過去了。慘絕人寰。

#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_JUDGEfreopen("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;
}

　　B：討論一發即可。

#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_JUDGEfreopen("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\n",ans);}return 0;
}

　　C：對矩形求個交，冷靜一下瞎算算就行了。

#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_JUDGEfreopen("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\n",white,black);}return 0;
}

　　D：考慮答案為i時的最小和最大分裂次數是多少，各種遞推式求通項到最后能搗鼓出來兩個式子。顯然答案若存在與n相差不會很大，暴力枚舉check即可。當然式子不能直接算，需要各種亂搞防溢出。數學太差推一年。

#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_JUDGEfreopen("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\n");else printf("YES %d\n",ans);}return 0;
}

　　E：這個條件比較顯然的等價于子矩形中每一行都能重排成回文串（即出現次數為奇數的字母不超過一個）且對稱行的字母組成相同。預處理每一行的每一段是否合法以及字母出現次數的哈希值，暴力枚舉子矩形左右端點，對上下端點的每段合法區間馬拉車即可。完全沒時間寫了。

　　F：沒看

　　小號打的。result：rank 139 rating +92

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