• K贪心


    <span style="color:#330099;">/*
    K - 贪心 基础
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
    Submit
     
    Status
    Description
    FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean. 
    The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain. 
     
    Input
    The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000. 
     
    Output
    For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain. 
     
    Sample Input
     5 3
    7 2
    4 3
    5 2
    20 3
    25 18
    24 15
    15 10
    -1 -1 
     
    Sample Output
     13.333
    31.500 
     BY Grant Yuan
     2014.7.14
     贪心
     */
     /*
    描写叙述
    有一个投资人。他有金钱m,可选投资项目n个。对于每一个投资项目,投入金钱c则获得收益f。要求对每一个项目所投入的资金r应该在0到c之间(即0<=r<=c),获得的收益为r*f/c。如今你的任务就是求出投资人能获得的最大收益。
    
    
    输入
    包含多个測例。每一个測例第一行为两个整数m,n,分别表示金钱数和项目个数。接下来n行,每行两个整数为收益f和投入c。输入最后以两个-1结尾。
    
    
    输出
    包含n行,每行为相应測例的最大收益。保留三位小数printf("%.3lf
    ",result);。
    
    
    输入例子
    3 3
    4 2
    2 1
    3 1
    4 3
    4 2
    5 3
    4 3
    -1 -1
    
    
    输出例子
    7.000
    7.333
    By yuan.c
    20146/22
    
    提示
    */
    #include<stdio.h>
    #include<iostream>
    #include<cstring>
    #include<cstdlib>
    using namespace std;
    int N;
    int M;
    
    double p[1002];
    int m[1002];
    int s[1002];
    double sum=0;
    
    void sort()
    {int t;double l;
        for(int i=0;i<N-1;i++)
           for(int j=i;j<N;j++)
             {
                 if(p[i]<p[j]){
                     l=p[i],p[i]=p[j],p[j]=l;
                     t=m[i],m[i]=m[j],m[j]=t;
                     t=s[i],s[i]=s[j],s[j]=t;
                 }
             }
    }
    
    int main()
    {
        while(1){
          scanf("%d %d",&M,&N);
          if(N==-1&&M==-1)
            break;
           for(int i=0;i<N;i++)
              scanf("%d %d",&s[i],&m[i]);
          for(int i=0;i<N;i++)
               p[i]=s[i]*1.0/m[i]*1.0;
         sort();
         for(int i=0;i<N;i++)
              if(M>=m[i])
                 sum+=s[i],M-=m[i];
            else {sum+=M*p[i];
            break;
            }
            printf("%.3lf
    ",sum);
            sum=0;
            memset(s,0,100);
            memset(m,0,100);
            memset(p,0,100);
        }
        return 0;
    }
    </span>

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  • 原文地址:https://www.cnblogs.com/lcchuguo/p/5236075.html
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