ACM 第三天
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit of 1378n.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
The single line of input contains one integer n (0 ≤ n ≤ 109).
Output
Print single integer — the last digit of 1378n.
Examples
1
8
2
4
Note
In the first example, last digit of 13781 = 1378 is 8.
In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
if(n==)
printf("1\n");
else if(n%==)
{
printf("8\n");
}
else if(n%==)
{
printf("4\n");
}
else if(n%==)
{
printf("2\n");
}
else
{
printf("6\n");
}
return ;
}
Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad.
Vladik knows n airports. All the airports are located on a straight line. Each airport has unique id from 1 to n, Vladik's house is situated next to the airport with id a, and the place of the olympiad is situated next to the airport with id b. It is possible that Vladik's house and the place of the olympiad are located near the same airport.
To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport a and finish it at the airport b.
Each airport belongs to one of two companies. The cost of flight from the airport i to the airport j is zero if both airports belong to the same company, and |i - j| if they belong to different companies.
Print the minimum cost Vladik has to pay to get to the olympiad.
The first line contains three integers n, a, and b (1 ≤ n ≤ 105, 1 ≤ a, b ≤ n) —
the number of airports, the id of the airport from which Vladik starts
his route and the id of the airport which he has to reach.
The second line contains a string with length n, which consists only of characters 0 and 1. If the i-th character in this string is 0, then i-th airport belongs to first company, otherwise it belongs to the second.
Output
Print single integer — the minimum cost Vladik has to pay to get to the olympiad.
Examples
4 1 4
1010
1
5 5 2
10110
0
Note
In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1.
In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company.
#include<stdio.h>
int main()
{
int n,a,b;
char s[];
scanf("%d %d %d",&n,&a,&b);
scanf("%s",s);
if(s[a-]==s[b-])
printf("0\n");
else
printf("1\n");
return ;
}
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (n - 1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1, 2, 1] after the first step, the sequence [1, 2, 1, 3, 1, 2, 1] after the second step.
The task is to find the value of the element with index k (the elements are numbered from 1) in the obtained sequence, i. e. after (n - 1) steps.
Please help Chloe to solve the problem!
The only line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2n - 1).
Output
Print single integer — the integer at the k-th position in the obtained sequence.
Examples
3 2
2
4 8
4
Note
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
#include<bits/stdc++.h>
using namespace std;
long long n,k;
int main()
{
scanf("%lld%lld",&n,&k);
cout<<log2(k&(-k))+<<endl;
}
Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form
.
Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.
If there is no such answer, print -1.
The single line contains single integer n (1 ≤ n ≤ 104).
Output
If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.
If there are multiple answers, print any of them.
Examples
3
2 7 42
7
7 8 56
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
if(n==)printf("-1\n");
else printf("%d %d %d\n",n,n+,n*(n+));
return ;
}
ACM 第三天的更多相关文章
- 【ACM】三点顺序
三点顺序 时间限制:1000 ms | 内存限制:65535 KB 难度:3 描述 现在给你不共线的三个点A,B,C的坐标,它们一定能组成一个三角形,现在让你判断A,B,C是顺时针给出的还是逆 ...
- 寒假的ACM训练三(PC110107/UVa10196)
#include <iostream> #include <string.h> using namespace std; char qp[10][10]; int result ...
- ACM第三次比赛 Big Chocolate
Problem G Big Chocolate Mohammad has recently visited Switzerland . As he loves his friends very muc ...
- ACM第三次比赛UVA11877 The Coco-Cola Store
Once upon a time, there is a special coco-cola store. If you return three empty bottles to the sho ...
- ACM第三题 完美立方
形如a3= b3 + c3 + d3的等式被称为完美立方等式.例如123= 63 + 83 + 103 .编写一个程序,对任给的正整数N (N≤100),寻找所有的四元组(a, b, c, d),使得 ...
- 今天的第一个程序-南阳acm输入三个数排序
#include<stdio.h>main(){ int a,b,c,t; scanf("%d%d%d",&a,&b,&c); ...
- (转)女生应该找一个玩ACM的男生
1.强烈的事业心 将来,他也一定会有自己热爱的事业.而且,男人最性感的时刻之一,就是他专心致志做事的时候.所以,找一个机会在他全神贯注玩ACM的时候,从侧面好好观察他,你就会发现我说的话没错. 2.永 ...
- ACM心情总结
已经快要12点了,然而还有5000字概率论论文没有动.在论文里,我本来是想要总结一下ACM竞赛中出现过的概率论题目,然而当敲打第一段前言的时候,我就迟疑了. 我问自己,ACM竞赛到底有什么现实意义. ...
- 【ACMER纷纷表示】女生应该找一个玩ACM的男生
1.强烈的事业心 将来,他也一定会有自己热爱的事业.而且,男人最性感的时刻之一,就是他专心致志做事的时候.所以,找一个机会在他全神贯注玩ACM的时候,从侧面好好观察他,你就会发现我说的话没错.2.永不 ...
随机推荐
- 关于windows下安装mysql数据库出现中文乱码的问题
首先需要在自己安装的mysql路径下新建一个my.ini文件,如下: 然后在my.ini文件中输入一下内容,主要控制编码问题的为红框部分,如下: 为了方便大家使用,可以复制以下代码: [WinMySQ ...
- day 28 黏包及黏包解决方案
1.缓冲区 每个socket被创建以后,都会分配两个缓冲区,输入缓冲区和输出缓冲区,默认大小都是8k,可以通过getsocket()获取,暂时存放传输数据,防止程序在发送的时候卡阻,提高代码运行效率. ...
- Python 爬虫 (三)
#对第一章的百度翻译封装的函数进行更新 1 from urllib import request, parse from urllib.error import HTTPError, URLError ...
- 单片机-C语言-定义和申明
以下代码是单片机程序,51单片机,编译器为HT-IDE3000, 简单来说 头文件中只能申明, 变量在头文件中申明时,要加上extern 这个关键字用来告诉编译器,变量在其它的文件中定义,为什么要在头 ...
- 深圳Uber优步司机奖励政策(12月28日到1月3日)
滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...
- 南京Uber优步司机奖励政策(7.20~7.26)
人民优步奖励前提 *必须满足当周平均评分4.5星及以上,且当周接单率70%及以上,满足以上所有前提即可获得当周奖励 *刷单和红线行为立即封号并取消当周全部奖励及车费! 滴滴快车单单2.5倍,注册地 ...
- jmeter关联三种常用方法
在LR中有自动关联跟手动关联,但在我看来手动关联更准确,在jmeter中,就只有手动关联 为什么要进行关联:对系统进行操作时,本次操作或下一次操作对服务器提交的请求,这参数里边有部分参数需要服务器返回 ...
- 第四模块:网络编程进阶&数据库开发 第1章·网络编程进阶
01-进程与程序的概念 02-操作系统介绍 03-操作系统发展历史-第一代计算机 04-操作系统发展历史-批处理系统 05-操作系统发展历史-多道技术 06-操作系统发展历史-分时操作系统 07-总结 ...
- Python安装教程最新版
Python安装教程最新版 目前Python官网已经更新到了最新版Python 3.7.1, 相比Python 2系列,它的兼容性不是太好, 不过应该会在不久的将来会全面解决.它的安装比较容易,具体步 ...
- TPO-12 C1 Revise a Hemingway paper
TPO-12 C1 Revise a Hemingway paper 第 1 段 1.Listen to a conversation between a student and a professo ...