A. Bank Robbery
time limit per test

2 seconds

memory limit per test  

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.

Input

The first line of input contains three space-separated integers, ab 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

Output a single integer: the maximum number of banknotes Oleg can take.

 
input
5 3 7
8
4 7 5 5 3 6 2 8
output
4
input
6 5 7
5
1 5 7 92 3
output
0
Note:

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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 【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“ ...

  6. 【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), ...

  7. 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 ...

  8. 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; ...

  9. 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

  1. python全栈开发_day4_if,while和for

    一.if 1)if的用途 if常用于判断. 2)if的语法 tag=True tag2=True if tag: print("代码") elif tag2: print(&quo ...

  2. 2018牛客多校1 - J Different Integers 莫队/主席树签到

    题意:给出n<5e4,a[1...n],单次1e5总次1e6次查询除去区间(L,R)的数的个数 开场5分钟:莫队是不可能莫队的,这道题是不可能莫队的 最后1小时:真香 具体操作没啥特别的,注意一 ...

  3. 第八次 Scrum Meeting

    第八次 Scrum Meeting 写在前面 会议时间 会议时长 会议地点 2019/4/12 22:00 20min 大运村1号楼6F 附Github仓库:WEDO 例会照片 工作情况总结(4.12 ...

  4. Oracle之索引

    简介 1.说明 1)索引是数据库对象之一,用于加快数据的检索,类似于书籍的索引.在数据库中索引可以减少数据库程序查询结果时需要读取的数据量,类似于在书籍中我们利用索引可以不用翻阅整本书即可找到想要的信 ...

  5. 初识iVew table表属性

    <template> <Table :row-class-name="rowClassName" :columns="columns1" :d ...

  6. RocketMQ 安装

    RocketMQ 安装 1.进入目录 cd /usr 2.下载 wget http://mirrors.tuna.tsinghua.edu.cn/apache/rocketmq/4.3.0/rocke ...

  7. 我的Python升级打怪之路【六】:面向对象(二)

    面向对象的一些相关知识点 一.isinstance(obj,cls) 检查实例obj是否是类cls的对象 class Foo(object): pass obj = Foo() isinstance( ...

  8. 利用C#结合net use命令破解域帐号密码

    背景 我的职业是程序猿,而所在的工作单位因各种原因,对上网帐号有严格控制,近期竟然把我们的上网帐号全部停用,作为程序猿,不能上网,就如同鱼儿没有水,煮饭没有米,必须想办法解决此问题.公司的局域网环境是 ...

  9. 转 shell中$(( )) 与 $( ) 还有${ }的区别

    原文 http://blog.zol.com.cn/2322/article_2321763.html $( ) 与 ` ` (反引号)在 bash shell 中,$( ) 与 ` ` (反引号) ...

  10. FocusBI: 使用Python爬虫为BI准备数据源(原创)

    关注微信公众号:FocusBI 查看更多文章:加QQ群:808774277 获取学习资料和一起探讨问题. <商业智能教程>pdf下载地址 链接:https://pan.baidu.com/ ...