PAT甲级真题 1070 Mooncake (25分) C++实现（同乙级1020，简单贪心法）

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本文介绍了一种计算在给定市场总需求下，如何通过合理选择不同种类月饼的销售数量来最大化利润的方法。考虑到月饼库存量和单价，通过算法确定最佳销售组合。

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题目

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

思路

同PAT乙级真题 1020 月饼 C++实现

代码

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

struct P{
    float num;
    float value;
    float price;
};

bool cmp(P &p1, P &p2){
    return p1.price > p2.price;
}

int main(){
    int n, d;
    scanf("%d %d", &n, &d);
    vector<P> p(n);
    for (int i=0; i<n; i++){
        scanf("%f", &p[i].num);
    }
    for (int i=0; i<n; i++){
        scanf("%f", &p[i].value);
        p[i].price = p[i].value / p[i].num;
    }
    sort(p.begin(), p.end(), cmp);
    float sum = 0.0;
    int i = 0;
    while (i<n && d>p[i].num){
        d -= p[i].num;
        sum += p[i].value;
        i++;
    }
    if (i<n){
        sum += (d * p[i].price);
    }
    printf("%.2f\n", sum);
    return 0;
}