Problem description

Linear Kingdom has exactly one tram line. It has n stops, numbered from 1 to n in the order of tram's movement. At the i-th stop ai passengers exit the tram, while bipassengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.

Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.

Input

The first line contains a single number n (2 ≤ n ≤ 1000) — the number of the tram's stops.

Then n lines follow, each contains two integers ai and bi (0 ≤ ai, bi ≤ 1000) — the number of passengers that exits the tram at the i-th stop, and the number of passengers that enter the tram at the i-th stop. The stops are given from the first to the last stop in the order of tram's movement.

  • The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that a1 = 0.
  • At the last stop, all the passengers exit the tram and it becomes empty. More formally, .
  • No passenger will enter the train at the last stop. That is, bn = 0.

Output

Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).

Examples

Input

4
0 3
2 5
4 2
4 0

Output

6

Note

For the first example, a capacity of 6 is sufficient:

  • At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3.
  • At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now.
  • At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now.
  • Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.

Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.

解题思路:题目的意思就是找出上下车过程中车内的最多人数,即为车的至少容量,简单水过。题目已经保证了一开始下车的人数为0,最后上车的人数也为0,并且最后车内的人将全部下车。

AC代码:

 #include<bits/stdc++.h>
using namespace std;
int main(){
int n,a,b,nowcap=,maxcap=;
cin>>n;
while(n--){
cin>>a>>b;
nowcap-=a;nowcap+=b;
maxcap=max(maxcap,nowcap);
}
cout<<maxcap<<endl;
return ;
}

C - Tram的更多相关文章

  1. [最短路径SPFA] POJ 1847 Tram

    Tram Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 14630 Accepted: 5397 Description Tra ...

  2. POJ 1847 Tram (最短路)

    Tram 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/N Description Tram network in Zagreb ...

  3. poj 1847 Tram【spfa最短路】

    Tram Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 12005   Accepted: 4365 Description ...

  4. (简单) POJ 1847 Tram,Dijkstra。

    Description Tram network in Zagreb consists of a number of intersections and rails connecting some o ...

  5. Codeforces Round #386 (Div. 2) C. Tram

    C. Tram time limit per test 1 second memory limit per test 256 megabytes input standard input output ...

  6. Tram

    Tram 题目大意:给你一个图,这个图上有n点和边.点上有开关,开关开始指向一条道路,拨动开关可以使开关指向由开关出发的任意一条路径,读入,a,b,求,至少要拨动几次才能从a点走到b点. 注释:n&l ...

  7. POJ1847 Tram

    Tram Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20274   Accepted: 7553 Description ...

  8. poj1847 Tram(最短路dijkstra)

    描述: Tram network in Zagreb consists of a number of intersections and rails connecting some of them. ...

  9. POJ 1847 Tram (最短路径)

    POJ 1847 Tram (最短路径) Description Tram network in Zagreb consists of a number of intersections and ra ...

  10. POJ1847:Tram(最短路)

    Tram Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20116   Accepted: 7491 题目链接:http:/ ...

随机推荐

  1. java中关于数组的初始化

  2. HDU 2414 Chessboard Dance(模拟题,仅此纪念我的堕落)

    题目 模拟题也各种wa,我最近真的堕落了,,,,,智商越来越为负数了!!!!!!!! #include<stdio.h> #include<string.h> #include ...

  3. colgroup 整行变色

    <table border="2" width="100%"> <colgroup span="2" align=&quo ...

  4. 18清明校内测试T3

    扫雷(mine) Time Limit:1000ms   Memory Limit:128MB 题目描述 rsy最近沉迷于一款叫扫雷的游戏. 这个游戏是这样的.一开始网格上有n*m个位置,其中有一些位 ...

  5. uva 12108 Extraordinarily Tired Students (UVA - 12108)

    算法完全转载...原博客(https://blog.csdn.net/u014800748/article/details/38407087) 题目简单叙述 题目就是一堆学生他们有清醒的时候和昏迷的时 ...

  6. hdu2002 计算球体积【C++】

    计算球体积 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submi ...

  7. crontab定时任务安装、使用方法

    本文介绍下,在linux中安装crontab的方法,以及crontab的具体用法,有需要的朋友参考下. 这里使用yum方式安装crontab:  复制代码代码示例: [root@CentOS ~]# ...

  8. noip模拟赛 括号序列

    题目描述LYK有一个括号序列,但这个序列不一定合法.一个合法的括号序列如下:()是合法的括号序列.若A是合法的括号序列,则(A)是合法的括号序列.若A和B分别是合法的括号序列,则AB是合法的括号序列. ...

  9. POJ - 3541 - Given a string…

    Given a string… Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 1819   Accepted: 390 C ...

  10. [luogu2209][USACO13]燃油经济性Fuel Economy_贪心

    燃油经济性Fuel Economy 题目大意:FJ想要去旅行.他的车总容量为G,每行驶一个单位就消耗一个单位的油.FJ要行驶D个单位的距离.期间存在n个加油站,每个加油站有一个价格,表示在这个燃油站买 ...