HDOJ 1009之FatMouse' Trade

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 66418 Accepted Submission(s): 22570

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 thatFatMouse can obtain.

Sample Input

5 3

7 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

大致题意：大肥鼠有m磅的猫粮，而大肥鼠最喜欢吃的JavaBean在仓库里，于是它要拿着猫粮去跟在仓库看守JavaBean的猫去换，仓库被分成n个部分。第i个部分的JavaBean有J[i]磅，需要F[i]磅猫粮兑换；大肥鼠并不用把该部分的全部JavaBean都兑换掉，而是按照每一部分的比例（权重）来兑换！你的任务就是算出大肥鼠能兑换JavaBean的最大值。

#include<stdio.h>#include<algorithm> using namespace std;struct node{    int j;    int f;    double w; //用w来表示各个部分的权重}a[1001]; bool cmp(node a,node b){return a.w>b.w;//自定义比较函数}int main(){int m,n;while(scanf("%d %d",&m,&n)&&(m!=-1||n!=-1)){//输入格式要求，当m，n都为-1是终止for(int i=0;i<n;i++){scanf("%d %d",&a[i].j,&a[i].f);a[i].w=a[i].j*1.0/a[i].f;//小技巧，int型转化为double型}sort(a,a+n,cmp);//使结构体按w从大到小排序double sum=0;//一定要定义在while循环内部！for(int i=0;i<n;i++){if(m>a[i].f){sum+=a[i].j;//如果大肥鼠拥有的猫粮M大于兑换所用的猫粮，则直接全部兑换m-=a[i].f;}else{sum+=a[i].w*m;//否则M乘以比值break;}}printf("%.3lf\n",sum);//输出格式要求}    return 0; }

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