【PAT甲级】1070 Mooncake（贪心）

Mooncake is a Chinese bakery product traditionally eaten during the Mid-Autumn Festival. Many types of fillings and crusts can be found in traditional mooncakes according to the region's culture. Now given the inventory amounts and the prices of all kinds of the mooncakes, together with the maximum total demand of the market, you are supposed to tell the maximum profit that can be made.

Note: partial inventory storage can be taken. The sample shows the following situation: given three kinds of mooncakes with inventory amounts being 180, 150, and 100 thousand tons, and the prices being 7.5, 7.2, and 4.5 billion yuans. If the market demand can be at most 200 thousand tons, the best we can do is to sell 150 thousand tons of the second kind of mooncake, and 50 thousand tons of the third kind. Hence the total profit is 7.2 + 4.5/2 = 9.45 (billion yuans).

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (≤1000), the number of different kinds of mooncakes, and D (≤500 thousand tons), the maximum total demand of the market. Then the second line gives the positive inventory amounts (in thousand tons), and the third line gives the positive prices (in billion yuans) of N kinds of mooncakes. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the maximum profit (in billion yuans) in one line, accurate up to 2 decimal places.

Sample Input:

3 200
180 150 100
7.5 7.2 4.5

Sample Output:

9.45

题目大意

这题就是给出月饼的种类数、需要卖出的、然后给出每种月饼存量和总售价。求出卖出需要卖出月饼的最大收益。

个人思路

这题是典型的贪心问题，只要求出每个月饼的平均利润（单价），然后从高到低排序，再卖到目标数量就可以啦。

【注：这题是乙级的1020】

代码实现

#include 
#include 
#include 
#include 
#include 

using namespace std;

int const maxn = 1005;

struct mooncake {
	double backup, price, profit;
};

mooncake mooncakes[maxn];

bool cmp(const mooncake m1, const mooncake m2) {
	return m1.profit > m2.profit;
}

int main() {
	//输入 
	int n, need;
	cin >> n >> need;
	for (int i = 0; i < n; i ++) {
		cin >> mooncakes[i].backup;
	}
	for (int i = 0; i < n; i ++) {
		cin >> mooncakes[i].price;
		mooncakes[i].profit = mooncakes[i].price / mooncakes[i].backup;
	}
	
	// 排序 
	sort(mooncakes, mooncakes+n, cmp);
	
	// 贪心 
	double sum = 0.0, ans = 0.0;
	for (int i = 0; i < n; i ++) {
		if (mooncakes[i].backup == 0) {
			continue;
		}
		// 这款月饼全卖了还不够
		if (sum + mooncakes[i].backup <= need) {
			sum += mooncakes[i].backup;
			ans += mooncakes[i].price;
		}
		// 这款月饼买一部分就可以啦
		else {
			ans += (need - sum) * mooncakes[i].profit;
			break;
		}
	}
	// 输出
	printf("%.2lf",ans);
	return 0;
}

总结

学习不息，继续加油