#HDOJ1009
#主要想通过这道题来练习下结构体快排函数的使用
#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.
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
#代码实现
#include<stdio.h> #include<stdlib.h> typedef struct fun_ { int J; int F; }fun; struct fun_ room[1000]; int comp(const void*a,const void *b) { double A,B; A=(double)(((fun *)a)->J)/(((fun *)a)->F); B=(double)(((fun *)b)->J)/(((fun *)b)->F); return(A>B?1:-1); } int main() { int M,N; while(scanf("%d%d",&M,&N)&&M!=-1) { double sum=0; //最终获得的数量 for(int i=0;i<N;i++) scanf("%d%d",&room[i].J,&room[i].F); qsort(room,N,sizeof(room[0]),comp); for(int i=N-1;i>=0;i--) { if(M>=room[i].F) //能获得房间内全部的 { sum+=room[i].J; M-=room[i].F; } else //只能获得部分的(按百分比折算) { sum+=M*(double)room[i].J/room[i].F; break; } } printf("%.3f\n",sum); } }#代码中的double不能换成float,否则AC不了,float只有6-7位有效数字。