A. Saitama Destroys Hotel
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 sand 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:s层楼 电梯1秒下一层  并且只能向下  n个人  在不同楼层 以及到达该楼层的时间

题解;到达的0层的时间

 #include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
using namespace std;
int n,s;
int a[];
int maxn;
int l,r;
int main()
{
scanf("%d %d",&n,&s);
for(int i=;i<=s;i++)
{
a[i]=;
}
for(int i=;i<=n;i++)
{
scanf("%d %d",&l,&r);
if(r>a[l])
a[l]=r;
}
int maxn=-;
for(int i=s;i>=;i--)
{
{
if(a[i]>maxn)
maxn=a[i];
else
maxn++;
}
}
cout<<maxn<<endl;
return ;
}

Codeforces Round #336 (Div. 2) A的更多相关文章

  1. 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 ...

  2. Codeforces Round #336 (Div. 2) D. Zuma 记忆化搜索

    D. Zuma 题目连接: http://www.codeforces.com/contest/608/problem/D Description Genos recently installed t ...

  3. Codeforces Round #336 (Div. 2) C. Chain Reaction set维护dp

    C. Chain Reaction 题目连接: http://www.codeforces.com/contest/608/problem/C Description There are n beac ...

  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)A. Saitama Destroys Hotel 水题

    A. Saitama Destroys Hotel 题目连接: http://www.codeforces.com/contest/608/problem/A Description Saitama ...

  6. Codeforces Round #336 (Div. 2)【A.思维,暴力,B.字符串,暴搜,前缀和,C.暴力,D,区间dp,E,字符串,数学】

    A. Saitama Destroys Hotel time limit per test:1 second memory limit per test:256 megabytes input:sta ...

  7. Codeforces Round #336 (Div. 2) D. Zuma(区间DP)

    题目链接:https://codeforces.com/contest/608/problem/D 题意:给出n个宝石的颜色ci,现在有一个操作,就是子串的颜色是回文串的区间可以通过一次操作消去,问最 ...

  8. Codeforces Round #336 (Div. 2)

    水 A - Saitama Destroys Hotel 简单的模拟,小贪心.其实只要求max (ans, t + f); #include <bits/stdc++.h> using n ...

  9. Codeforces Round #336 (Div. 2) D. Zuma 区间dp

    D. Zuma   Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gems ...

  10. Codeforces Round #336 (Div. 2)C. Chain Reaction DP

    C. Chain Reaction   There are n beacons located at distinct positions on a number line. The i-th bea ...

随机推荐

  1. 【yii2】 yii框架如果控制器和方法都是多个单词组成应该怎样写请求链接

    最近的一个项目碰到这种问题,摸索出了解决办法,故此记录下 example 如果控制器为 ShopCollect 方法为 UserList solution 请求的链接应该为 doman.com/sho ...

  2. linux无名管道

    特点 无名管道是半双工的,也就是说,一个管道要么只能读,要么只能写 只能在有共同祖先的进程间使用(父子进程.兄弟进程.子孙进程等) fork或者execve调用创建的子进程,继承了父进程的文件描述符 ...

  3. POJ:2456-Aggressive cows

    Aggressive cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 18313 Accepted: 8716 Desc ...

  4. 十、mysql之索引原理与慢查询优化

    mysql之索引原理与慢查询优化 一.介绍 1.什么是索引? 一般的应用系统,读写比例在10:1左右,而且插入操作和一般的更新操作很少出现性能问题,在生产环境中,我们遇到最多的,也是最容易出问题的,还 ...

  5. POJ 3461 Oulipo(字符串hash)

    题目链接 字符串hash判断字符串是否相等. code #include<cstdio> #include<algorithm> #include<cstring> ...

  6. (C)spring boot读取自定义配置文件时乱码解决办法

    这是入门的第三天了,从简单的hello spring开始,已经慢慢接近web的样子.接下来当然是读取简单的对象属性了. 于是按照网上各位大神教的,简单写了个对象book,如上一篇(B),其他配置不需要 ...

  7. python考点

    Python考点 1.Python类继承,内存管理(阿里) 答:内存管理机制包括:引用计数机制,垃圾回收机制,内存池机制 a = 1,1就是对象,a就是引用,引用a指向对象1. 2.Python装饰器 ...

  8. CentOS-6.3-minimal安装gnome桌面环境(转载)

    最近,想学着搞搞linux,从入门安装开始,先装centos6.3-minimal,发现是windowser最不习惯的命令界面,先升级桌面,教程如下. 1.添加一个普通用户,并设置密码useradd  ...

  9. 剑指Offer - 九度1368 - 二叉树中和为某一值的路径

    剑指Offer - 九度1368 - 二叉树中和为某一值的路径2013-11-23 03:46 题目描述: 输入一颗二叉树和一个整数,打印出二叉树中结点值的和为输入整数的所有路径.路径定义为从树的根结 ...

  10. 《Cracking the Coding Interview》——第17章:普通题——题目12

    2014-04-29 00:04 题目:给定一个整数数组,找出所有加起来为指定和的数对. 解法1:可以用哈希表保存数组元素,做到O(n)时间的算法. 代码: // 17.12 Given an arr ...