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PAT B1020 -《算法笔记》

作者:互联网

前言:

21考研,正在啃《算法笔记》,不论能否进复试记录一下准备路上写下的垃圾代码。

解答:

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;

struct moon {
	double save;
	double price;
	double portion;
};
bool cmp(moon brand1, moon brand2) {
	if (brand1.portion > brand2.portion)
		return true;
	else
		return false;
}
int main() {
	int n, req;
	double income = 0;
	cin >> n >> req;
	vector<moon> brand;
	moon temp;
	for (int i = 0; i < n; i++)
		brand.push_back(temp);
	for (int i = 0; i < n; i++)
		cin >> brand[i].save;
	for (int i = 0; i < n; i++)
		cin >> brand[i].price;
	for (int i = 0; i < n; i++)
		brand[i].portion = brand[i].price / brand[i].save;
	sort(brand.begin(), brand.end(), cmp);	//先卖高性价比的

	double sum_save = 0;		
	for (int i = 0; i < n; i++)
		sum_save += brand[i].save;
	if (sum_save >= req) {			//测试点3,出现库存小于需求
		for (int i = 0; req >= 0; i++) {
			if (req >= brand[i].save) {	
				req -= brand[i].save;
				income += brand[i].price;
			}
			else {
				income += brand[i].price / brand[i].save * req;
				req = -1;
			}
		}
	}
	else {
		for (int i = 0; i < n; i++)
			income += brand[i].price;
	}
	printf("%.2lf\n", income);
	return 0;
}

标签:PAT,int,brand,req,算法,++,B1020,save,price
来源: https://blog.csdn.net/weixin_44897291/article/details/113523296