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
- laravel5.8的使用
首先,确定电脑已经安装了composer.最好是全局安装 然后打开phpstorm的控制台: composer create-project --prefer-dist laravel/laravel ...
- Python yield 的基本概念和用法
之前解析MQTT协议时,需要做一个等分字节流的操作,其中用到了yield关键字,如下: def get_var_length(hstring): m = 1 v = 0 for element in ...
- EXCEL 使用宏 打水印
Sub 宏1() ' ' 宏1 宏 //用于调整列宽 加边框 ' ' 快捷键: Ctrl+w ' ActiveCell.Cells.Select ActiveCell.Cells.EntireColu ...
- 关于:Mac下IDEA更换Maven仓库
一.MAC安装配置maven环境变量 1.下载maven包: 下载链接:
- 移远EC20的使用
一 发短信 3. 推荐短信流程3.1 查询 短信存储区AT+CPMS?+CPMS: "ME",19,255,"ME",19,255,"ME" ...
- openerp学习笔记 对象间关系【多对一(一对一)、一对多(主细结构)、多对多关系、自关联关系(树状结构)】
1.多对一(一对一)关系:采购单与供应商之间的关系 'partner_id':fields.many2one('res.partner', 'Supplier', required=True, sta ...
- MySQL 的数据库、表基本操作
1.链接数据库 mysql -u root -ppassword 2创建数据库 create database mr_book; 3选择数据库 use mr_book; 4 创建表 create ta ...
- greenplum表的distributed key值查看
greenplum属于分布式的数据库,MPP+Share nothing的体系,查询的效率很快.不过,这是建立在数据分散均匀的基础上的.如果DK值设置不合理的话,完全有可能出现所有数据落在单个节点上的 ...
- sql developer链接不上oracle 报 The Network Adapter could not establish the connection
安装时候报 : Oracle 支持在具有 DHCP 分配的公共 IP 地址的系统上进行安装.但应使用静态 IP 地址来配置系统的主网络接口, 以便 Oracle 软件正常工作.有关在配置有 DHCP ...
- android studio应用获取系统属性权限(SystemProperties)
dependencies { provided files(getLayoutLibPath()) } /** ZhangChao time:2014-12-31,get layoutlib.jar ...