CodeForces999A-Mishka and Contest
1 second
256 megabytes
standard input
standard output
Mishka started participating in a programming contest. There are nn problems in the contest. Mishka's problem-solving skill is equal to kk.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than kk. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by 11. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
The first line of input contains two integers nn and kk (1≤n,k≤1001≤n,k≤100) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100), where aiai is the difficulty of the ii-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Print one integer — the maximum number of problems Mishka can solve.
8 4
4 2 3 1 5 1 6 4
5
5 2
3 1 2 1 3
0
5 100
12 34 55 43 21
5
In the first example, Mishka can solve problems in the following order: [4,2,3,1,5,1,6,4]→[2,3,1,5,1,6,4]→[2,3,1,5,1,6]→[3,1,5,1,6]→[1,5,1,6]→[5,1,6][4,2,3,1,5,1,6,4]→[2,3,1,5,1,6,4]→[2,3,1,5,1,6]→[3,1,5,1,6]→[1,5,1,6]→[5,1,6], so the number of solved problems will be equal to 55.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than kk.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
ac代码;
#include<bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n,k,a[110],flag1,flag2,cnt=0;
cin>>n>>k;
flag1=1,flag2=n;
for(int i=1;i<=n;i++) cin>>a[i];
while(flag1<=flag2)
{
while(a[flag1]<=k && flag1<=flag2)
{
cnt++;
flag1++;
}
while(a[flag2]<=k && flag1<=flag2)
{
cnt++;
flag2--;
}
if(a[flag1]>k&&a[flag2]>k) break;
}
cout<<cnt<<endl;
return 0;
}
CodeForces999A-Mishka and Contest的更多相关文章
- Mishka and Contest(模拟水题)
Mishka started participating in a programming contest. There are nn problems in the contest. Mishka' ...
- CF A.Mishka and Contest【双指针/模拟】
[链接]:CF/4892 [题意]: 一个人解决n个问题,这个问题的值比k小, 每次只能解决最左边的或者最右边的问题 解决了就消失了.问这个人能解决多少个问题. [代码]: #include<b ...
- CF999A Mishka and Contest 题解
Content 能力值为 \(k\) 的小 M 参加一次考试,考试一共有 \(n\) 道题目,每道题目的难度为 \(a_i\).小 M 会选择两头中的一道难度不超过他的能力值题目去做,每做完一道,这道 ...
- ACM 第二天
A - Mishka and Contest Mishka started participating in a programming contest. There are n problems i ...
- [Codeforces]Codeforces Round #490 (Div. 3)
Mishka and Contest #pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JU ...
- Codeforces Round #365 (Div. 2) D. Mishka and Interesting sum 离线+线段树
题目链接: http://codeforces.com/contest/703/problem/D D. Mishka and Interesting sum time limit per test ...
- Codeforces 703B. Mishka and trip 模拟
B. Mishka and trip time limit per test:1 second memory limit per test:256 megabytes input:standard i ...
- CodeForces-999A-Mishka and Contest
Mishka started participating in a programming contest. There are nn problems in the contest. Mishka' ...
- Programming Contest Problem Types
Programming Contest Problem Types Hal Burch conducted an analysis over spring break of 1999 and ...
随机推荐
- PowerMock学习(一)之PoweMock的入门--模拟新增学生操作
关于powermock 在TDD领域Mock框架有很多,比如EasyMock,JMock,Mockito.可能有些同学会好奇了,为什么要重点把powermock拿出来呢,因为powermock可以解决 ...
- maven聚合(依赖聚合)
maven聚合工程 原文地址:http://juvenshun.iteye.com/blog/305865 http://blog.csdn.NET/woxueliuyun/article/detai ...
- Flsk&pyecharts 动态数据可视化
1:数据源 Hollywood Movie Dataset: 好莱坞2006-2011数据集 实验目的: 实现 统计2006-2011的数据综合统计情况,进行数据可视化 gitee地址: https ...
- 使用boost实现线程池thread pool | boost thread pool example
本文首发于个人博客https://kezunlin.me/post/f241bd30/,欢迎阅读! boost thread pool example Guide boost thread pool ...
- Mysql 添加字段 修改字段 删除字段
1.添加字段(alter.add) mysql> alter table users add name varchar(30) not null after id; 2.修改字段(alter.m ...
- Liunx 安装配置zsh和oh-my-zsh 替换 bash
一.前言 本文将基于 Liunx 环境安装配置zsh 和 oh-my-zsh 替换 bash oh my zsh Liunx默认shell是单调的bash,而zsh比较高大上,bash有的功能,zsh ...
- 【Luogu P3375】字符串匹配KMP算法模板
Luogu P3375 模式串:即题目中的S2所代表的意义 文本串:即题目中的S1所代表的意义 对于字符串匹配,有一种很显然的朴素算法:在S1中枚举起点一位一位匹配,失配之后起点往后移动一位,从头开始 ...
- python 面向对象的基本概念(未完待续)
面向对象编程简称OOP(Object-oriented-programming),是一种程序设计思想. 面向过程编程(如C语言)指一件事该怎么做,面向对象编程(如Java.python)指一件事该让谁 ...
- 利用堆来处理Top K问题
目录 一.什么是Top K问题 二.Top K的实际应用场景 三.Top K问题的代码实现及其效率对比 1.用堆来实现Top K 2.用快排来实现Top K 3.用堆或用快排来实现 TopK 的效率对 ...
- CentOS 7上利用systemctl添加自定义系统服务
Centos 7 之 systemctl CentOS 7继承了RHEL 7的新的特性,例如强大的systemctl,而systemctl的使用也使得以往系统服务的/etc/init.d的启动脚本的方 ...