time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to s and elevator initially starts on floor s at time 0.

The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.

Input

The first line of input contains two integers n and s (1 ≤ n ≤ 100, 1 ≤ s ≤ 1000) — the number of passengers and the number of the top floor respectively.

The next n lines each contain two space-separated integers fi and ti (1 ≤ fi ≤ s, 1 ≤ ti ≤ 1000) — the floor and the time of arrival in seconds for the passenger number i.

Output

Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.

Examples
Input
3 7
2 1
3 8
5 2
Output
11
Input
5 10
2 77
3 33
8 21
9 12
10 64
Output
79
Note

In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:

1. Move to floor 5: takes 2 seconds.

2. Pick up passenger 3.

3. Move to floor 3: takes 2 seconds.

4. Wait for passenger 2 to arrive: takes 4 seconds.

5. Pick up passenger 2.

6. Go to floor 2: takes 1 second.

7. Pick up passenger 1.

8. Go to floor 0: takes 2 seconds.

This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds.

题意就是电梯接乘客,这个电梯只能从上往下走,给的数据是在第几层有乘客,乘客到达该层的时间,乘客上电梯的时间忽略不计,要求输出花费的最短时间是多少。

我的理解就是单独把某层的时间拿出来算一下,从该层到最底层花的时间+等待乘客的时间和总的层数花的时间比较一下,找出max,然后其他层也是这样,和上一组楼层的max比较一下。其实总的想一下,就是把楼层数+等待乘客的时间找出max,然后和总的楼层数比较一下就可以了。(最近好多题都用到这种思想了,不知道这种思想叫啥,但是感觉有时候挺好用的,就是有时用这种思路写题会T,超时了耶。。。)

代码:

#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m,f,t;
int maxx=-;
while(~scanf("%d%d",&n,&m)){
for(int i=;i<n;i++){
scanf("%d%d",&f,&t);
if(f+t>maxx)
maxx=max(f+t,maxx);
}
if(maxx<m)
printf("%d\n",m);
else
printf("%d\n",maxx);
}
return ;
}
 

Codeforces Round #336 (Div. 2)-608A.水题 608B.前缀和的更多相关文章

  1. A. Yellow Cards ( Codeforces Round #585 (Div. 2) 思维水题

    ---恢复内容开始--- output standard output The final match of the Berland Football Cup has been held recent ...

  2. Codeforces Round #345(Div. 2)-651A.水题 651B.。。。 651C.去重操作 真是让人头大

    A. Joysticks time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  3. Codeforces Round #315 (Div. 2) B 水题强行set

    B. Inventory time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  4. Codeforces Round #336 (Div. 2)B. Hamming Distance Sum 前缀和

    B. Hamming Distance Sum 题目连接: http://www.codeforces.com/contest/608/problem/A Description Genos need ...

  5. Codeforces Round #336 (Div. 2) D. Zuma

    Codeforces Round #336 (Div. 2) D. Zuma 题意:输入一个字符串:每次消去一个回文串,问最少消去的次数为多少? 思路:一般对于可以从中间操作的,一般看成是从头开始(因 ...

  6. Codeforces Round #609 (Div. 2)前五题题解

    Codeforces Round #609 (Div. 2)前五题题解 补题补题…… C题写挂了好几个次,最后一题看了好久题解才懂……我太迟钝了…… 然后因为longlong调了半个小时…… A.Eq ...

  7. Codeforces Round #297 (Div. 2)B. Pasha and String 前缀和

    Codeforces Round #297 (Div. 2)B. Pasha and String Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx ...

  8. Educational Codeforces Round 7 B. The Time 水题

    B. The Time 题目连接: http://www.codeforces.com/contest/622/problem/B Description You are given the curr ...

  9. Educational Codeforces Round 7 A. Infinite Sequence 水题

    A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/622/problem/A Description Consider the ...

随机推荐

  1. iOS 让图片变模糊

    #import <Accelerate/Accelerate.h> 1.初始化图片 UIImageView *iv = [[UIImageView alloc]initWithFrame: ...

  2. java随机生成验证码

    package com.yuyuchen.util; import java.awt.Color; import java.awt.Font; import java.awt.Graphics; im ...

  3. 学习 Kubernetes 的 Why 和 How - 每天5分钟玩转 Docker 容器技术(114)

    这是一个系统学习 Kubernetes 的教程,有下面两个特点: 系统讲解当前最流行的容器编排引擎 Kubernetes包括了安装部署.应用管理.网络.存储.监控.日志管理等多各个方面. 重实践并兼顾 ...

  4. 3.Nginx常用功能介绍

    Nginx常用功能介绍 Nginx反向代理应用实例 反向代理(Reverse Proxy)方式是指通过代理服务器来接受Internet上的连接请求,然后将请求转发给内部网络上的服务器,并且从内部网络服 ...

  5. 串口调试者v2.1------开源c#串口调试工具

    第一步:上图 第二步:上代码 >>>>>>>>>>>源代码下载<<<<<<<<< ...

  6. Java NIO (一) 初识NIO

    Java NIO(New IO / Non-Blocking IO)是从JDK 1.4版本开始引入的IO API , 可以替代标准的Java IO API .NIO与原来标准IO有同样的作用和目的,但 ...

  7. RedisPool操作Redis,工具类实例

    redis.properties 配置文件内容 redis.pool.maxActive=100redis.pool.maxIdle=20redis.pool.maxWait=3000redis.po ...

  8. SecureCRT连接本地的Vmware虚拟机(CentOS)时提示连接超时“Connection timed out”

    测试了一下,直接在Vmware的VM里面可以ping通宿主机. 但是宿主机无法ping通VM. 后面发现是本地的网络设置里面的vmware的NAT的网卡设置了手工填写地址和DNS. 修改为自动获取.问 ...

  9. 微信小程序+“芝麻小客服”可设自动关注公众号,助力运营闭环

    微信小程序全面上线已经接近1年的时间,从最初的"用完即走"理念到2017年总计更新开放60余次的功能创新,微信小程序不一定会爆发下一次的红利,但绝对是微信生态中重要的一环. 芝麻小 ...

  10. MIG IP控制DDR3读写测试

    本文设计思想采用明德扬至简设计法.在高速信号处理场合下,很短时间内就要缓存大量的数据,这时片内存储资源已经远远不够了.DDR SDRAM因其极高的性价比几乎是每一款中高档FPGA开发板的首选外部存储芯 ...