传送门:
http://acm.hdu.edu.cn/showproblem.php?pid=1009
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 93676 Accepted Submission(s): 32566
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 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
Recommend
题目意思:
一共有n个房子,每个房子里有老鼠喜欢吃的javabeans,但是每个房间里的javabeans的价格不一样。老鼠用m元,问m元最多可以卖多少javabeans,其中每个房间里的javabeans可以被分割。
分析:按照单价排序(价值除以重量),一直选择单价最大的的全部,当不能选择全部的时候,选择一部分
code:
#include<iostream> #include<algorithm> #include<stdio.h> #include<queue> #include<set> #include<map> #include<string> #include<memory.h> #include<math.h> #define eps 1e-7 using namespace std; #define max_v 1005 struct node { double v,c; }p[max_v]; bool cmp(node a,node b) { return (a.v/a.c)>(b.v/b.c);//单价 } int main() { int n,m; double sum; int i; while(cin>>m>>n) { if(n==-1&&m==-1) break; for(i=0;i<n;i++) cin>>p[i].v>>p[i].c; sort(p,p+n,cmp); sum=0; for(i=0;i<n;i++) { if(m>=p[i].c) { sum+=p[i].v; m-=p[i].c; }else { sum+=(m*(p[i].v/p[i].c)); break; } } printf("%0.3lf ",sum); } return 0; }