杭电oj 1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76080    Accepted Submission(s): 26061

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

13.333

31.500

刚写完的时候，测试数据都过了，但是提交就是不过，数据类型转换也没错啊，也都*1.0了，但是就是不对，最后把变量的类型都定义为double,结果就对了.....醉了。。

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;struct node{    double j;   //之前定义为int   double  f;    //之前定义为int    double avg;}s[1015];int m,n;double sum;bool cmp(node a,node b){    if(a.avg>b.avg)        return true;    else        return false;}int main(){    while(scanf("%d%d",&m,&n)!=EOF)    {        sum=0;        memset(s,0,sizeof(s));        if(m==-1&&n==-1)            break;        else        {                for(int i=0;i<n;i++)            {                scanf("%lf%lf",&s[i].j,&s[i].f);                s[i].avg=s[i].j*1.0/(s[i].f*1.0);            }            sort(s,s+n,cmp);                       for(int i=0;m!=0;i++)            {                if(m>=s[i].f)                {                    m=m-s[i].f;                    sum=sum+s[i].j;                }                else                {                    sum=sum+s[i].j*(m/(s[i].f*1.0));                    m=0;                }            }                printf("%0.3lf\n",sum);        }                    }    return 0;}