杭电1009

来源:互联网 发布:数据分析师薪酬待遇 编辑:程序博客网 时间:2024/06/16 05:21

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62414    Accepted Submission(s): 21050


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 <stdio.h>



typedef struct node
{
    int j;
    int f;
    double radio;
}Node;


int main(){
    Node list[1005];
    int cat,n;
    while(scanf("%d %d",&cat,&n)!=EOF && cat != -1 && n != -1){
        int i = 0;
        for(i = 0; i < n; i++){
            scanf("%d %d",&list[i].j,&list[i].f);
            list[i].radio = (double)list[i].j/list[i].f;
            if(i){
                Node key = list[i];
                //printf("key is %d %d %lf\n",key.j,key.f,key.radio);
                int j = i - 1;
                while(j >=0 && key.radio > list[j].radio){
                    list[j + 1] = list[j];
                    j--;
                }
                list[j + 1] = key;
            }
        }
       /* for(i = 0; i < n; i++){
            printf("%lf\n",list[i].radio);
        }*/
        double sum;
        for(i = 0, sum = 0; cat && i < n; i++){
            if(list[i].f > cat){
                sum += cat * list[i].radio;
                cat = 0;
            }else{
                cat -= list[i].f;
                sum += list[i].j;
            }
        }
        printf("%.3lf\n",sum );
    }
    return 0;
}
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