IsTheOrderARabbit??

NC222153. IsTheOrderARabbit??

描述

In a remote place, there is a rabbits market where rabbits can be bought and sold in every day. The trading rules in this market are as follows:

1. The market is available only in morning and afternoon in one day. And the rabbit transaction price may be different every day even different between morning and afternoon in a day.
2. It is prohibited that purchasing more than once in a day. And only one rabbit can be purchased at a time.
3. You can only sell once in a day, but you can sell rabbits unlimited quantities (provided that you have so many rabbits)

Mr.Cocktail has traded in this market for ${N}$ days, suppose he has unlimited money to buy rabbits. Before the first day and after the last day, he has no rabbits. Now, in these ${N}$ days, how much money did he earn the most?

输入描述

The first line contains a positive integer , which means there are  days in total .

Next, there are  lines, each line contains two positive integers, representing the rabbit price in the morning and afternoon of the day .

输出描述

Output an integer on a line to indicate the answer.

示例1

输入:

输出:

说明:

In example 1, a rabbit was purchased for $1 in the morning of the first day, and a rabbit was purchased for $2 in the afternoon of the second day. On the morning of the third day, the two rabbits were sold, a total profit of $11.

示例2

输入:

2
5 4
3 2

输出:

说明:

The price of the rabbit in sample 2 continues to decrease, and it is impossible to make money through buying and selling, so choose not to make any buying and selling, and output the answer 0.

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

C++ 解法, 执行用时: 81ms, 内存消耗: 1912K, 提交时间: 2021-05-22 16:00:27

#include <iostream>
using namespace std;
int main ()
{
	long long n,i,sum,max,a[100010],b[100010];
	cin>>n;
	for(i=1;i<=n;i++) cin>>a[i]>>b[i];
	max=b[n];
	for(i=n;i>0;i--)
	{
		if(b[i]>=max)
		{
			max=b[i];
			if(a[i]>=max) max=a[i];
			else sum=sum+max-a[i];
		}
		else if (a[i]>=max)
		{
			sum=sum+max-b[i];
			max=a[i];
		}
		else
		{
			if(a[i]>=b[i]) sum=sum+max-b[i];
			else sum=sum+max-a[i];
		}
	}
	cout<<sum;
	return 0;
}

上一题

下一题

详情