HDU 1009 FatMouse' Trade

题目

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

HDU 1009
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

代码

先给出我的AC代码，然后一段一段代码讲
AC代码：

#include<stdio.h>
int main(){
    int a[1001],b[1001];
    int i,m,n,j,temp;
    double sum,x;
    while(scanf("%d%d",&m,&n),m!=-1&&n!=-1){
        for(i=0;i<n;i++)
        scanf("%d%d",&a[i],&b[i]);
        for(i=0;i<n;i++)
        for(j=0;j<n-1;j++){
            if(a[j]*b[j+1]<a[j+1]*b[j]){
                temp=a[j];
                a[j]=a[j+1];
                a[j+1]=temp;
                temp=b[j];
                b[j]=b[j+1];
                b[j+1]=temp;
            }
        }
        sum=0;
        for(i=0;i<n;i++){
            if(b[i]>m){
            x=(double)m*a[i]/b[i];
            sum+=x;
            break;
            }
            if(b[i]<=m) {           
            sum+=a[i];
            m-=b[i];
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}

题解

题目大意是：一个人手里有m个cat food，然后有n个房间
其中每个房间有a[i]个JavaBeans ，和换取所对应需要的b[j]个cat food，非常需要注意的是假如手中的cat food足够时，必须交足该房间所需的cat food数，不足时按比例获得JavaBeans 数。
我的想法是通过比例的大小来满足贪心的要求，先通过比例将输入的a[i]与b[j],从新排序，使其新顺序满足从下标为0到下标结束时，单位换取的JavaBeans 量是最多的。
下面解剖代码：

for(i=0;i<n;i++)
        scanf("%d%d",&a[i],&b[i]);

接受多组房间的JavaBeans数 ，和换取所对应需要的cat food数。

        for(j=0;j<n-1;j++){
            if(a[j]*b[j+1]<a[j+1]*b[j]){
                temp=a[j];
                a[j]=a[j+1];
                a[j+1]=temp;
                temp=b[j];
                b[j]=b[j+1];
                b[j+1]=temp;
            }
        }

按照比例大小进行排序，temp是中间值，满足两个数组交换值，注意！！a[]与b[]一定要同时换，因为a[]与b[]是一一对应的。

 if(b[i]>m){
            x=(double)m*a[i]/b[i];
            sum+=x;
            break;
            }

该段代码是在手里的cat food不足的情况下时，能换取的JavaBeans 数