第一发博客

       今天是第一天集训开始，心情是激动的，也是非常开心。。学长先是讲了指针，，实际上没明白多少还是得靠自学。以前以为自学的人都很流弊，到了大学才明白全是靠自学的。也是有点小开心，，我想我应该能坚持到最后，不放弃。吉首大学的生活感觉还是挺好的。。在这次集训中我要尽量多学知识。。既然选了这门专业就该好好学。对没错是这样的。来发题目玩玩杭电1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46454    Accepted Submission(s): 15580

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

#include<cstdio>#include<iostream>#include<algorithm>using namespace std;struct p{    int mao;    int cost;    double aver;}num[1005];bool cmp(p x,p y){    return x.aver>y.aver;}int main(){    int money,n;    while(~scanf("%d %d",&money,&n))    {        if(money==-1)break;        double sum=0;        for(int i=0;i<n;i++)        {            scanf("%d %d",&num[i].mao,&num[i].cost);            num[i].aver=num[i].mao*1.0/num[i].cost;        }        sort(num,num+n,cmp);        for(int i=0;i<n;i++)        {            if(money<=num[i].cost)            {                sum+=money*num[i].aver;                break;            }            else            {                sum+=num[i].mao;                money-=num[i].cost;            }        }        printf("%.3f\n",sum);    }    return 0;}