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[PAT-A] 1070 Mooncake(25)

題目連結 - 技術文件

題目描述
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
題目分析
題目大意:給定不同種類月餅的數量和價格,現在市場需求一定數量的月餅,那麼如何保證收益最大 解題思路:貪心演算法.求出每種月餅單價,先售出單價最高的月餅,這樣一來保證了收益最高,但需要注意的是,題目未說月餅總數是整數,因為單位也不是以個數開始的
程式碼設計
//AC程式碼
//zhicheng
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
// zhicheng
// Octobrer,9,2018
typedef struct{double num;double
prices;}D; D d[1005]; bool comp(D a,D b){return (a.prices/a.num)>(b.prices/b.num);} int main() { int n;double k,re=0;scanf("%d %lf",&n,&k); for(int i=0;i<n;i++) scanf("%lf",&d[i].num); for(int i=0;i<n;i++) scanf("%lf",&d[i].prices); sort(d,d+n,comp); for(int i=
0;k&&i<n;i++) { if(k>=d[i].num) {k-=d[i].num;re+=d[i].prices;} else {re+=(d[i].prices/d[i].num)*k;break;} } printf("%.2lf\n",re); return 0; }

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