A. Arrays(Codeforces Round #317 水题)
2 seconds
256 megabytes
standard input
standard output
You are given two arrays A and B consisting
of integers, sorted in non-decreasing order. Check whether it is possible to choose knumbers
in array A and choose m numbers
in array B so that any number chosen in the first array is strictly less than any number chosen in the second array.
The first line contains two integers nA, nB (1 ≤ nA, nB ≤ 105),
separated by a space — the sizes of arrays A and B,
correspondingly.
The second line contains two integers k and m (1 ≤ k ≤ nA, 1 ≤ m ≤ nB),
separated by a space.
The third line contains nA numbers a1, a2, ... anA ( - 109 ≤ a1 ≤ a2 ≤ ... ≤ anA ≤ 109),
separated by spaces — elements of array A.
The fourth line contains nB integers b1, b2, ... bnB ( - 109 ≤ b1 ≤ b2 ≤ ... ≤ bnB ≤ 109),
separated by spaces — elements of array B.
Print "YES" (without the quotes), if you can choose k numbers
in array A and m numbers
in array B so that any number chosen in arrayA was
strictly less than any number chosen in array B. Otherwise, print "NO"
(without the quotes).
3 3
2 1
1 2 3
3 4 5
YES
3 3
3 3
1 2 3
3 4 5
NO
5 2
3 1
1 1 1 1 1
2 2
YES
In the first sample test you can, for example, choose numbers 1 and 2 from array A and number 3 from array B (1
< 3 and 2 < 3).
In the second sample test the only way to choose k elements in the first array and m elements
in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in A will be less than all the numbers
chosen in B:
.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm> using namespace std;
int a[100100],b[100010];
int n,m;
int k,t; int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
scanf("%d%d",&k,&t);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
for(int i=0; i<m; i++)
{
scanf("%d",&b[i]);
}
if(a[k-1] < b[m-t])
{
printf("YES\n");
}
else
{
printf("NO\n");
} }
return 0;
}
A. Arrays(Codeforces Round #317 水题)的更多相关文章
- Coprime Arrays CodeForces - 915G (数论水题)
反演一下可以得到$b_i=\sum\limits_{d=1}^i{\mu(i)(\lfloor \frac{i}{d} \rfloor})^n$ 整除分块的话会T, 可以维护一个差分, 优化到$O(n ...
- codeforces 659A A. Round House(水题)
题目链接: A. Round House time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Educational Codeforces Round 27 补题
题目链接:http://codeforces.com/contest/845 A. Chess Tourney 水题,排序之后判断第n个元素和n+1个元素是不是想等就可以了. #include < ...
- Codeforces数据结构(水题)小结
最近在使用codeblock,所以就先刷一些水题上上手 使用codeblock遇到的问题 1.无法进行编译-------从setting中的编译器设置中配置编译器 2.建立cpp后无法调试------ ...
- Educational Codeforces Round 15 套题
这套题最后一题不会,然后先放一下,最后一题应该是大数据结构题 A:求连续最长严格递增的的串,O(n)简单dp #include <cstdio> #include <cstdlib& ...
- CodeForces 705A Hulk (水题)
题意:输入一个 n,让你输出一行字符串. 析:很水题,只要判定奇偶性,输出就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,10240 ...
- Codeforces Round #346 (Div. 2) A. Round House 水题
A. Round House 题目连接: http://www.codeforces.com/contest/659/problem/A Description Vasya lives in a ro ...
- Educational Codeforces Round 21 A-E题题解
A题 ............太水就不说了,贴下代码 #include<string> #include<iostream> #include<cstring& ...
- Codeforces Round #317 (div 2)
Problem A Arrays 思路:水一水. #include<bits/stdc++.h> using namespace std; ; int n1,n2,k,m,a[N],b[N ...
随机推荐
- 呵呵哒,LNMP下通过fread方式下载文件时,中文名称文件找不到文件
哎,整整折腾一个下午. 本来好好的,thinkphp 自动的uniq方式保存的文件名,非要使用原文件名,真心蛋疼~~ 然后就只好写个脚本 把原来的所有文件都重新命名一下 - - 然后把数据库对应字段也 ...
- Python学习笔记整理总结【语言基础篇】
一.变量赋值及命名规则① 声明一个变量及赋值 #!/usr/bin/env python # -*- coding:utf-8 -*- # _author_soloLi name1="sol ...
- hadoop的webUI查看Live Nodes为1
开起了两个节点,而且jps查看确实开启了,但是用web端50070查看却一直显示为1 经过排查,将虚拟机直接copy一份,但是之前配置好hadoop环境的namenode格式化(format)生成的文 ...
- CentOS6.9编译安装Nginx1.12
1:安装必要的库 Bash yum install gc gcc gcc-c++ pcre-devel zlib-devel openssl-devel 2:创建Nginx用户和组 Bash grou ...
- 使用Microsoft.AspNetCore.TestHost进行完整的功能测试
简介 Microsoft.AspNetCore.TestHost是可以用于Asp.net Core 的功能测试工具.很多时候我们一个接口写好了,单元测试什么的也都ok了,需要完整调试一下,检查下单元测 ...
- ELK 快速指南
ELK 快速指南 概念 ELK 是什么 ELK 是 elastic 公司旗下三款产品 ElasticSearch .Logstash .Kibana 的首字母组合. ElasticSearch 是一个 ...
- Pyqt5学习系列
最近在学习Pyqt5做界面,找到了一个非常棒的博主的学习系列 在此记录下来: http://blog.csdn.net/zhulove86/article/category/6381941
- 用原型代替PRD时,原型应该包含哪些内容
随着互联网节奏越来越快,传统的需求文档已经比较难适应市场的脚步,特别对于要求敏捷的团队来说,冗余而细致入微的需求文档已经成为包袱(这么长个文档领导也不会看呀).目前大多数团队更喜爱直接使用原型来代替需 ...
- 【smart-transform】取自 Atom 的 babeljs/coffeescript/typescript 智能转 es5 库
简介 有时间研究下开源库的源码,总是会有些收获的.注意到 Atom 插件编写时,可以直接使用 babel, coffeescript 或者 typescript.有些诧异,毕竟 Electron 中内 ...
- MYSQL常见运算符和函数
字符函数 (1)CONCAT():字符连接 SELECT CONCAT('IMOOC','-','MySQL');//IMOOC-MySQL SELECT CONCAT (first_name,las ...