Codeforces Round #414 A. Bank Robbery
2 seconds
256 megabytes
A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes.
Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the i-th safe from the left is called safe i. There are n banknotes left in all the safes in total. The i-th banknote is in safe xi. Oleg is now at safe a. There are two security guards, one of which guards the safe b such that b < a, i.e. the first guard is to the left of Oleg. The other guard guards the safe c so that c > a, i.e. he is to the right of Oleg.
The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.
The first line of input contains three space-separated integers, a, b and c (1 ≤ b < a < c ≤ 109), denoting the positions of Oleg, the first security guard and the second security guard, respectively.
The next line of input contains a single integer n (1 ≤ n ≤ 105), denoting the number of banknotes.
The next line of input contains n space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 109), denoting that the i-th banknote is located in thexi-th safe. Note that xi are not guaranteed to be distinct.
Output a single integer: the maximum number of banknotes Oleg can take.
5 3 7
8
4 7 5 5 3 6 2 8
4
6 5 7
5
1 5 7 92 3
0
In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes.
For the second sample, Oleg can't take any banknotes without bumping into any of the security guards.
题目大意:
有一个小偷想去偷银行的钱,他打听到银行一共有n个保险箱在一条直线上给出每个保险箱的位置
(每一个位置上不一定只有一个保险箱)
小偷现在在a,有两个保安他们分别在b和c处(a > b a < c) 两个保安特别懒他们保持不动
小偷不能越过保安去偷钱,问小偷可以偷多少个保险箱
解题思路:
可能是时间长不敲代码了吧。被这么水的题困了半天。
输入一个位置只要判断当前位置是否在两个保安的位置之间,是+1 然后输出即可(真是被自己蠢到了)
AC代码:
#include<stdio.h>
int main ()
{
int a,b,c,i,n,p,sum;
while (~scanf("%d%d%d",&a,&b,&c))
{
scanf("%d",&n); sum = ;
for (i = ; i < n; i ++)
{
scanf("%d",&p);
if (p > b && p < c)
sum ++;
}
printf("%d\n",sum);
}
return ;
}
Codeforces Round #414 A. Bank Robbery的更多相关文章
- Tinkoff Challenge - Final Round (Codeforces Round #414, rated, Div. 1 + Div. 2) 【ABC】
老年人题解,语言python3 A - Bank Robbery 题意:给你ABC,以及n个数,问你在(B,C)之间的数有多少个. 题解:对于每个数判断一下就好了嘛 x,y,z = map(int,i ...
- Tinkoff Challenge - Final Round (Codeforces Round #414, rated, Div. 1 + Div. 2) 继续跪一把
这次的前三题挺简单的,可是我做的不快也不对. A. Bank Robbery time limit per test 2 seconds memory limit per test 256 megab ...
- Codeforces Round #414 C. Naming Company
http://codeforces.com/contest/794/problem/C 题意: 有两个人要为公司起名字,每个人手中都有n个字符,现在要取一个n个字符长度的公司名.两人轮流取名,每次选择 ...
- Tinkoff Challenge - Final Round (Codeforces Round #414, rated, Div. 1 + Div. 2)
A: 思路:就是找b,c之前有多个s[i] 代码: #include<stdio.h>#define ll long longusing namespace std;ll a,b,c;in ...
- 【构造】Tinkoff Challenge - Final Round (Codeforces Round #414, rated, Div. 1 + Div. 2) D. Labelling Cities
考试的时候想的是,将所有的完全子图缩起来,然后如果剩下的是一条链,依次对其进行标号即可. 看了官方题解,发现完全子图这个条件太强了,缩点的条件仅仅需要保证原本两个点的“邻接表”相同即可.(注意这里的“ ...
- 【贪心】【multiset】Tinkoff Challenge - Final Round (Codeforces Round #414, rated, Div. 1 + Div. 2) C. Naming Company
考虑两个人,先把各自的集合排个序,丢掉一半,因为比较劣的那一半一定用不到. 然后贪心地放,只有两种决策,要么把一个最优的放在开头,要么把一个最劣的放在结尾. 如果我的最优的比对方所有的都劣(或等于), ...
- Codeforces Round #414
A =w= B qvq C(贪心) 题意: Alice和Bob分别有长度为n(n<=1e5)的字符串,Alice先手,每次从自己的字符串中抽出一个字母放到新字符串的某个位置,一共轮流n次,也就是 ...
- codeforces round #414 div1+div2
A:判断一下就可以了 #include<bits/stdc++.h> using namespace std; typedef long long ll; int a, b, c, n; ...
- Codeforces Round #366 (Div. 2) ABC
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...
随机推荐
- 洛谷 P3285 / loj 2212 [SCOI2014] 方伯伯的 OJ 题解【平衡树】【线段树】
平衡树分裂钛好玩辣! 题目描述 方伯伯正在做他的 OJ.现在他在处理 OJ 上的用户排名问题. OJ 上注册了 \(n\) 个用户,编号为 \(1\sim n\),一开始他们按照编号排名.方伯伯会按照 ...
- 使用maven搭建ssm框架的javaweb项目
目前主流的javaweb项目,常会用到ssm(Spring+Spring MVC+Mybatis)框架来搭建项目的主体框架,本篇介绍搭建SSM框架的maven项目的实施流程.记之共享! 一.SSM框架 ...
- ajax 的error参数
ajax发生错误会进入到error中,我们在这里把错误信息从控制台中输出出来,为了避免每次写ajax都得写好几个console.log(). 我在这里写一个模板,需要用的时候直接过来拷贝. error ...
- 使用AChartEngine画动态曲线图
AChartEngine是一个开源的Android图表库,可以用来画折线图.平滑折线图.饼图.直方图等等.使用简单,功能强大. AChartEngine官网:http://www.achartengi ...
- JS模块加载系统设计V1
一.require模块 +function() { var path = location.protocol + "//" + location.host +"/Java ...
- iOS设备的屏幕分辨率
全部列在这里吧.方便自己方便别人.保持更新…… iPhone: iPhone 1G320x480 iPhone 3G320x480 iPhone 3GS320x480 iPhone 4640x960 ...
- 遇到Caused by: java.lang.NoClassDefFoundError: javax/validation/ParameterNameProvider
今天在做spring和hibernate整合的时候遇到这个问题 网上搜找到这里有解决办法 http://blog.csdn.net/jueshengtianya/article/details/122 ...
- Jquery jqXHR对象的属性和方法
在 jQuery 1.4 之前(包括1.4),$.ajax() 方法返回的是浏览器原生的 XMLHttpRequest 对象. 从 jQuery 1.5 开始,$.ajax() 方法返回 jQuery ...
- Sed - An Introduction and Tutorial by Bruce Barnett
http://www.grymoire.com/unix/sed.html Quick Links - NEW Sed Commands : label # comment {....} Block ...
- Vue父子组件生命周期执行顺序及钩子函数的个人理解
先附一张官网上的vue实例的生命周期图,每个Vue实例在被创建的时候都需要经过一系列的初始化过程,例如需要设置数据监听,编译模板,将实例挂载到DOM并在数据变化时更新DOM等.同时在这个过程中也会运行 ...