HDU 1009 FatMouse' Trade

FatMouse' Trade

http://acm.hdu.edu.cn/showproblem.php?pid=1009

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28122    Accepted Submission(s): 9028

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

 

 

sort版本

 

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>

using namespace std;

struct node{
    int j,f;
    double r;
}food[1005];

int cmp(node a,node b){
    return a.r>b.r;
}

int main(){
    int m,n;
    while(scanf("%d%d",&m,&n)){
        if(m==-1 && n==-1)
            break;
        int i;
        for(i=0;i<n;i++){
            scanf("%d%d",&food[i].j,&food[i].f);
            food[i].r=(double)food[i].j/food[i].f;
        }
        //qsort(food,n,sizeof(node),cmp);
        sort(food,food+n,cmp);
        double ans=0;
        for(i=0;i<n;i++)
            if(m>=food[i].f){
                m-=food[i].f;
                ans+=food[i].j;
            }else{
                ans+=(food[i].j*1.0/food[i].f)*m;
                break;
            }
        printf("%.3lf\n",ans);
    }
    return 0;
}

qsort版本

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>

using namespace std;

struct node{
    int j,f;
    double r;
}food[1005];

int cmp(const void *a,const void *b){
    node c=*(node *)a;
    node d=*(node *)b;
    if(c.r>d.r)
        return -1;
    else
        return 1;
}

int main(){
    int m,n;
    while(scanf("%d%d",&m,&n)){
        if(m==-1 && n==-1)
            break;
        int i;
        for(i=0;i<n;i++){
            scanf("%d%d",&food[i].j,&food[i].f);
            food[i].r=(double)food[i].j/food[i].f;
        }
        qsort(food,n,sizeof(food[0]),cmp);
        //sort(food,food+n,cmp);
        double ans=0;
        for(i=0;i<n;i++)
            if(m>=food[i].f){
                m-=food[i].f;
                ans+=food[i].j;
            }else{
                ans+=(food[i].j*1.0/food[i].f)*m;
                break;
            }
        printf("%.3lf\n",ans);
    }
    return 0;
}