Codeforces刷题之路——116A Tram

本文介绍了一种计算公交线路车辆所需最小容量的方法。通过记录每站上下车人数,模拟乘客流动情况,确保车辆容量满足需求的同时尽可能减少浪费。
A. Tram
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 aipassengers exit the tram, while bi passengers 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.


题目大意:模拟一个公交车上下乘客的场景,起始站不下人,终点站不上人,输出该条线路上公交车能够满足乘客的最小容量数值。
题目思路:这是一道水题,采用一个数组存放每一站上下车完毕后车上乘客的数量,采用Arrays.sort(),自动排序该数组,以递增序列,所以只需要输出最后一项即可。


以下为解题代码(java实现)
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] passenger = new int[n+1];
		passenger[0] = 0;
		
		int a = 0;
		int b = 0;
		for(int i = 1; i <= n;i++){
			a = scanner.nextInt();
			b = scanner.nextInt();
			passenger[i] = (passenger[i-1] - a) + b;
		}
		Arrays.sort(passenger);
		System.out.println(passenger[passenger.length-1]);
	}
}
### Codeforces 编程练习 Codeforces 是一个广受欢迎的在线编程竞赛平台,提供了大量不同难度级别的目供用户练习。对于希望提升算法能力的学习者来说,该平台是一个非常有价值的资源。 #### 数学类目 一些数学类型的目可以帮助提高逻辑思维能力和解决问的能力。例如,在优快云博客上有一篇关于Codeforces简单数学的文章进行了详细的总结[^1]。这些目通常涉及数论、组合数学和其他基础数学概念的应用。 #### 字符串变换问 另一个有趣的类别是字符串处理目。比如在Educational Codeforces Round 39 中有一个名为 "String Transformation" 的挑战[C. String Transformation][^2]。这类目往往考察选手对字符操作的理解以及如何高效地转换给定条件下的字符串序列。 #### 提升技巧的方法 为了更好地利用Codeforces来增强解技能并达到更高的评级水平,可以从以下几个方面入手: - **克服弱点**:通过分析个人比赛表现找出薄弱环节(如动态规划、图论等问),针对性加强训练。 - **构建工具库**:创建常用数据结构和算法模板(像区间最值查询(RMQ),树状数组(BIT), 线段树等),这有助于加快编码速度减少错误率[^3]。 ```cpp // 这里给出一段简单的RMQ实现作为例子 #include <vector> using namespace std; class RMQ { vector<int> rmq; public: void build(const vector<int>& nums){ int n = nums.size(); rmq.resize(n * 4); construct(nums, 0, n - 1, 1); } private: void construct(const vector<int>& nums,int l ,int r ,int node){ if(l == r){ rmq[node]=nums[l]; return ; } int mid=(l+r)/2; construct(nums,l,mid,node*2); construct(nums,mid+1,r,node*2+1); rmq[node]=min(rmq[node*2],rmq[node*2+1]); } }; ```
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