cf div2 #449 C.Nephren gives a riddle

Badcase归因分析“四部曲” badcase是业内的专业名词,简单解释一下。bad是坏,case是案例,说白了就是模型或者具体规则无法很好预测的那些案例。以分类模型为例,本来一个样本应该是负类的,结果预测为正类,那这个样本就能被称为bad case。那相对应的就是good case就是优秀案例。 阅读详情

Nephren gives a riddle

题意:给你一个初始串f[0]:"What are you doing at the end of the world? Are you busy? Will you save us?"(不包括引号)。为了让这句话被更多人知道定义串f[i]为"What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?"(不包括第一个与最后一个引号),接下来有q次询问,每次询问第n个串的第k个字符是什么,这个字符可能是字母(大小写都有可能)、空格、问号或引号(因为每次引用前一个串的时候都会加上一个引号)。


思路:用一个数组来存放每个串的长度,而长度很好获得,从对串f[i]的定义中可以很容易的得到:从第一个串开始,每个串可以分为五个部分,其中:

第二、第四个串为f[i-1];

第一个串为:What are you doing while sending "(对没错,包括引号,这样才能保证第二、第四个串为f[i-1]);

第三个串为:"? Are you busy? Will you send ";

第五个串为:"?

其中第一、三、五个串的长度可以得到分别是:34、32、2,而第二、第四个串可以调用f[i-1]。

比赛的时候以为询问时k最大1e18就能用long long直接存下,wa的莫名其妙,赛后看了别人代码的 n >= 54 恍然大悟:当n为53时,串的长度已经达到1.2*1e18,所以53之后的串的前1e18个字符和第53个串的差别只有最前面的第一个串,所以对n和k同时处理即可。



AC代码:

/***********************************************
	Author:	StupidTurtle
	Date:		2017.12.05
	Email:		31601359@std.zucc.edu.cn 
***********************************************/ 

#include <iostream>
#include <cstring>
#include <string>

using namespace std ;

typedef long long ll ;

ll q , n , k ;
string Head , Mid , End , F0 ;

ll l[100005] ;

void solve ( ll n , ll k ){
	if ( n != 0 ){
		if ( k <= 34 ){
			cout << Head[k-1] ;
			return ;
		} else if ( k <= 34 + l[n-1] ){
			solve ( n-1 , k-34 ) ;
		} else if ( k <= 66 + l[n-1] ){
			cout << Mid[k-34-l[n-1]-1] ;
			return ;
		} else if ( k <= 66 + l[n-1] * 2 ){
			solve ( n-1 , k-66-l[n-1] );
		} else {
			cout << End[k-66-l[n-1]*2-1] ;
			return ;
		}
	} else {
		cout << F0[k-1] ;
	}
}

int main(void){

	Head = "What are you doing while sending \"" ;
//	cout << Head.length() << endl ;
	Mid = "\"? Are you busy? Will you send \"" ;
//	cout << Mid.length() << endl ;
	End = "\"?" ;
//	cout << End.length() << endl ;
	F0 = "What are you doing at the end of the world? Are you busy? Will you save us?" ;
//	cout << F0.length() << endl ;
	l[0] = 75 ;
	for ( int i = 1 ; i <= 100000 ; i ++ ){
		if ( i >= 54 )  l[i] = l[i-1];
    else    l[i] = l[i-1] * 2 + 68 ;
	}
	cin >> q ;
	for ( int i = 0 ; i < q ; i ++ ){
		cin >> n >> k ;
    while ( n > 53 && k > 34 ){
        n -- ;
        k -= 34 ;
    }
		if ( k > l[n] )		cout << '.' ;
		else	solve ( n , k );
	}
	cout << endl ;
	
	return 0;
}


c#算法(5)—最小二乘法(二次曲线拟合) 最小二乘法——二次曲线拟合 阅读详情

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传送门题意:Petr和Alex两个人对一个初始为1,2,3,...,n的长度为n的排列有不同的操作次数(一次操作表现为将其中两个数交换位置,比如1,2,3,4变为1,2,4,3称为一次操作),Petr会对其进行3n次操作,而Alex会对其进行7n+1次操作。现在给你初始排列的长度n和结果排列,问你是他们中的谁进行的操作,如果是Petr输出"Petr",否则输出"Um_nik"。思路:利用到排列的性...

Stupid_Turtle的博客 1097

Codeforces 981D Bookshelves(贪心+dp)

传送门题意:给你n本书以及每本书的权值,现在让你把n本书放到k个书架上(只有连续的几本书可以放到一个书架上),每个书架的权值是书架上每本书的权值加和,总的"beauty"是每个书架权值按位与的结果,要求输出最大的"beauty"。思路:因为要求总"beauty"是所有书架权值的按位与的结果,所以对于总"beauty"的二进制,位数越高的1的价值越大,所以可以从高位开始枚举这一位是否能取1,如果能取...

Stupid_Turtle的博客 906

2019, XII Samara Regional Intercollegiate Programming Contest 解题报告

2019, XII Samara Regional Intercollegiate Programming Contest 传送门 A. Rooms and Passages 题意: 有 n+1n+1n+1 个房间,nnn 把钥匙,有第 iii 把钥匙你可以从房间 i−1i-1i−1 走到房间 iii。每个房间 iii 都有两种类型中的一种: 一、只有拥有钥匙 aia_iai​ 才能从 i−1...

Stupid_Turtle的博客 828

2018, XI Samara Regional Intercollegiate Programming Contest 解题报告

2018, XI Samara Regional Intercollegiate Programming Contest 传送门 A. Restoring Numbers 题意: 给出两个数 aaa 和 bbb 的和 sss 和最大公因数 ggg,要你求出任意一组 aaa 和 bbb 的解。 题解: 我们可以知道 gcd(a,b)=ggcd(a,b)=ggcd(a,b)=g,则我们假设 a=p∗...

Stupid_Turtle的博客 814

codeforces 1080D Olya and magical square (思维+数学)(模拟)

传送门:codeforces 1080D 题意:给你一个 n 和 k ,表示你现在有一个边长为 2^n 的正方形,你需要对其进行恰好 k 次操作,使得:① 左下角和右上角的正方形边长一样,假设为 a;② 存在一条路径,这条路径连通左下角和右上角的正方形,并且这条路径上的所有正方形边长也都为 a。其中,操作指的是:找到目前图形中一个边长不为 1 的正方形,假设边长为 b,把它变成四个边长为 b/2...

Stupid_Turtle的博客 769

codeforces 1017D The Wu(状压+预处理)

传送门:codeforces 1017D 题意:给你n m q,表示在这一组数据中所有的01串长度均为n,然后给你一个含有m个元素的multiset,之后有q次询问。每次询问会给你一个01串t和一个给定常数k,让你输出串t和multiset里面多少个元素的“Wu”值不超过k。对于“Wu”值的定义:如果两个01串s和t在位置i上满足s[i]==t[i],那么加上w[i],处理完s和t的所有n位之后...

Stupid_Turtle的博客 739

Codeforces 985E Pencils and Boxes(尺取法/two pointers)

传送门题意:给定n,k,d,表示给你n支铅笔,每支有一个权值v。现在让你把n支笔放入一些盒子中(盒子数量可以无穷大),每个盒子中至少有k支笔,而且每个盒子中的笔的max(v)-min(v)不超过d。问你能否找到一个合法的放法,可以输出"YES",否则输出"NO"。思路:因为对权值差有要求,所以先进行排序,排序后能放进一个盒子的笔的权值v一定是连续的。之后记录两个值,一个是can[i],表示第i支笔...

Stupid_Turtle的博客 663

cf div2 #453 ABC(D)

赛中只搞出ABC,C想的还有点久,没时间搞D了(虽然是推了很久都没推出来),蒟蒻的上分之旅。 A. Visiting a Friend 题意: 你的初始位置为0,给你n和m,m表示你的目的地,n表示可乘坐工具的个数。接下来n行,每行一个x y,表示在位置x处有一个工具,你可以坐着它到位置y(你可以中途跳车,比如样例1中,0->2,2->4,在2->4过程中跳车,在位置3坐上3->

Stupid_Turtle的博客 545
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