CodeForces-116A-Tram

本文介绍了一个简单的算法,用于计算线性王国中唯一一条电车线路的最小容量需求,确保在任何时候乘客数量都不会超过该容量。通过输入每个站点上下车的乘客数量,算法能够有效地找出电车所需的最小容量。

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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 ai passengers 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).

Sample test(s)
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.

import java.util.*;

public class Tram {
	public static void main(String[] args) {
		Scanner inScanner = new Scanner(System.in);
		int n = inScanner.nextInt();
		
		int mintotal = 0, temp = 0;//ini

		while (n > 0) {
			temp-=inScanner.nextInt();//update
			temp+=inScanner.nextInt();//update
			if(temp>mintotal)
				mintotal=temp;
			n--;//update
		}
		System.out.println(mintotal);
	}

}
改进的思路,在对最大值进行判断时,可以使用类函数Math.max,以节省代码量。即mintotal=Math.max(temp,mintotal);

### 关于 Codeforces Problem 1802A 目前提供的引用内容并未涉及 Codeforces 编号为 1802A 的题目详情或解决方案[^1]。然而,基于常见的竞赛编程问题模式以及可能的解决方法,可以推测该类题目通常围绕算法设计、数据结构应用或者特定技巧展开。 如果假设此题属于典型的算法挑战之一,则可以从以下几个方面入手分析: #### 可能的方向一:字符串处理 许多入门级到中级难度的问题会考察字符串操作能力。例如判断子串是否存在、统计字符频率或是执行某种转换逻辑等。以下是 Python 中实现的一个简单例子用于演示如何高效地比较两个字符串是否相匹配: ```python def are_strings_equal(s1, s2): if len(s1) != len(s2): return False for i in range(len(s1)): if s1[i] != s2[i]: return False return True ``` #### 方向二:数组与列表的操作 另一常见主题是对整数序列进行各种形式上的变换或者是查询最值等问题。下面给出一段 C++ 程序片段来展示快速寻找最大元素位置的方法: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ int n; cin >> n; vector<int> a(n); for(auto &x : a){ cin>>x; } auto max_it = max_element(a.begin(),a.end()); cout << distance(a.begin(),max_it)+1; // 输出索引加一作为答案 } ``` 由于具体描述缺失,在这里仅提供通用框架供参考。对于确切解答还需要访问实际页面获取更多信息后再做进一步探讨[^3]。
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