1070. Mooncake (25)-PAT甲级
来源:互联网 发布:vb编程入门 编辑:程序博客网 时间:2024/05/22 00:20
题目:
同B1020:
http://blog.csdn.net/zzlcsdn2017/article/details/66819465
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
解答:
以前竟然用float,好傻,以后坚决不用,因为float精度特别低,稍微多加运算就会导致答案错误。
#include<cstdio>#include<algorithm>using namespace std;struct mooncake{ double store; double sell; double price;}cake[1010];bool cmp(mooncake a,mooncake b){ return a.price>b.price; }int main(){ int n; double D; scanf("%d%lf",&n,&D); for(int i=0;i<n;i++){ scanf("%lf",&cake[i].store); } for(int i=0;i<n;i++){ scanf("%lf",&cake[i].sell); cake[i].price=cake[i].sell/cake[i].store; } sort(cake,cake+n,cmp); double ans=0; for(int i=0;i<n;i++){ if(cake[i].store<=D){ D-=cake[i].store; ans+=cake[i].sell; }else{ ans+=cake[i].price*D; break;//不要忘了 } } printf("%.2f",ans); return 0;}
- PAT甲级1070. Mooncake (25)
- 1070. Mooncake (25)-PAT甲级
- 【PAT甲级】1070. Mooncake (25)
- PAT 甲级 1070. Mooncake (25)
- 1070. Mooncake (25)-PAT甲级真题
- PAT甲级练习1070. Mooncake (25)
- 1020 月饼 PAT PAT乙级&&1070. Mooncake (25) PAT甲级
- pat甲级 10700---Mooncake
- 【PAT】1070. Mooncake (25)
- PAT 1070. Mooncake (25)
- PAT 1070. Mooncake (25)
- PAT 1070. Mooncake (25)
- pat 1070. Mooncake (25)
- PAT 1070. Mooncake (25)
- PAT 1070. Mooncake (25)
- PAT (Advanced) 1070. Mooncake (25)
- PAT A 1070. Mooncake (25)
- ZJU-PAT 1070. Mooncake (25)
- life is cool
- 数据库三范式
- mysql数据库修改密码
- Servlet新特性
- easy ui一些控件常见赋值取值方法
- 1070. Mooncake (25)-PAT甲级
- 梯度下降法
- 晚安日志
- 基于android的网络音乐播放器-网络音乐的多线程下载(六)
- Java中使用SimpleEmail实现简单邮件的发送
- oracle11g空表不能导出
- 在servlet中对IP进行限制
- webstrom 设置禁用拼写检查
- 【转载】浏览器进化史