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    上题目

    FatMouse' Trade

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 33316    Accepted Submission(s): 10813


    Problem 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
     
     
          题意就是有m单位猫食,n堆的豆,每堆豆有一个余量和要这一堆豆的代价,每一堆豆可以只要一部分,那么付出的代价就按这一部分占这一堆豆的余量的百分比算。问用这m单位猫食可以最多得到多少豆。
        贪心的基础题,白书上也有类似的题目。解法就是得出每一堆豆的单价(余量/代价),先取完单价最高的豆,然后再取第二的···直到猫食用完。
     
    上代码:
     1 #include <stdio.h>
     2 #include <string.h>
     3 #include <algorithm>
     4 #define MAX 1000+10
     5 using namespace std;
     6 
     7 typedef struct
     8 {
     9     double J,F;
    10     double S;
    11 }price;
    12 
    13 bool cmp(price x,price y) {return x.S>y.S;}
    14 
    15 price p[MAX];
    16 
    17 int main()
    18 {
    19     int i,m,n;
    20     double sum;
    21     //freopen("data.txt","r",stdin);
    22     while(scanf("%d %d",&m,&n)!=EOF)
    23     {
    24         if(m==-1 && n==-1) break;
    25         memset(p,0,sizeof(p));
    26         for(i=0;i<n;i++)
    27         {
    28             scanf("%lf %lf",&p[i].J,&p[i].F);
    29             p[i].S=p[i].J/p[i].F;
    30         }
    31         sort(p,p+n,cmp);
    32         sum=0;
    33         for(i=0;i<n;i++)
    34         {
    35             if(m>=p[i].F) {m-=p[i].F;sum+=p[i].J;}
    36             else {sum+=p[i].J*m/p[i].F;break;}
    37         }
    38         printf("%.3lf
    ",sum);
    39     }
    40     return 0;
    41 }
    1009
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  • 原文地址:https://www.cnblogs.com/sineatos/p/3203090.html
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