zoukankan      html  css  js  c++  java
  • Pick apples 第三届acm省赛

    Description

    Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she has never seen so many apples before. Each kind of apple has a size and a price to be sold. Now the little girl wants to gain more profits, but she does not know how. So she asks you for help, and tell she the most profits she can gain.

    Input

    In the first line there is an integer T (T <= 50), indicates the number of test cases.
    In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S <= 100) and the price (1 <= P <= 10000) of this kind of apple.

    In the fourth line there is an integer V,(1 <= V <= 100,000,000)indicates the volume of the girl's bag.

    Output

    For each case, first output the case number then follow the most profits she can gain.

    Sample Input

    1
    1 1
    2 1
    3 1
    6
    

    Sample Output

    Case 1: 6

    #include <iostream>
    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    struct N
    {
        int s,p;
        double pri;
    } node[4];
    bool cmp(N a,N b)
    {
        if(a.pri<b.pri)        return true;
        return false;
    }
    long long llmax(long long a,long long b)
    {
        return a>b?a:b;
    }
    int main()
    {
        int t;
        cin>>t;
        int c=1;
        while(t--)
        {
            for(int i=0;i<3;i++){
                cin>>node[i].s>>node[i].p;
                node[i].pri = 1.0*node[i].s/(1.0*node[i].p);
            }
            int V;
            cin>>V;
            sort(node,node+3,cmp);
            long long ans=0;
            for(int i=0; i<node[0].s; i++)
            {
                for(int j=0; j<node[0].s; j++)
                {
                    long long temp=i*node[1].s+j*node[2].s;
                    if(temp>V)
                        break;
                    else
                    {
                        long long v=V-temp;
                        ans=llmax(ans,v/node[0].s*node[0].p+i*node[1].p+j*node[2].p);
                    }
                }
            }
            cout<<"Case "<<c++<<": "<<ans<<endl;
        }
        return 0;
    }
  • 相关阅读:
    [LeetCode]Surrounded Regions
    生产者消费者问题
    多线程试题汇总
    Linux多线程编程
    运算符优先级表
    正向代理和反向代理
    遗传算法
    嵌入式培训学习历程第十五天
    嵌入式培训学习历程第十四天
    一个猜数的游戏
  • 原文地址:https://www.cnblogs.com/upstart/p/6783571.html
Copyright © 2011-2022 走看看