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FatMouse‘ Trade
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 69 Accepted Submission(s) : 18
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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
Author
CHEN, Yue
Source
ZJCPC2004
这题的描述主要是：老鼠有m镑猫粮准备和猫换javebean。现在有n间房子，里面各有若干javebean可供老鼠兑换，兑换的条件是每个房间有各自所需的猫粮。例如5 3 7 2 4 3 5 2这组数据表示老鼠有5镑的猫粮，共有3间房子，每间房子依次有7,4,5镑的javebean各需要2,3,2猫粮换。我们要求的是用5镑换最多的javebean。我们可以知道用2镑换第一间房子的javebean用2镑换第3间房子的javebean，最后用1镑换第二间房子（1*（4/3）的javebean，一共13.333。
AC代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
struct Fun{
int CatFood,JavaBean;
double ratio;
};
Fun t[10000],temp;
bool c(Fun a,Fun b){
return a.ratio>b.ratio;
}
int main(){
int m,n,i,j;
while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)){
i=0;
while(i<n){
cin>>t[i].JavaBean>>t[i].CatFood;
t[i].ratio=(double)t[i].JavaBean/t[i++].CatFood;
}
sort(t,t+n,c);
double sum=0;
for(i=0;i<n&&m!=0;i++){
if(m>=t[i].CatFood){
sum+=t[i].JavaBean;
m-=t[i].CatFood;
}
else {
sum+=m*((double)t[i].JavaBean/t[i++].CatFood);
break;
}
}
printf("%.3f\n",sum);
}
return 0;
}
这里用结构体来储存数据，然后调用sort（）；进行排序计算出最优解。//因为有个参数写错导致一直WA，我找了半天没找到orz，细心啊！sort用法sort（n，n+m，comp）；n是数组名，m是数组长度，comp是自定义函数按照函数规定来进行排序。
运行结果：
HDU FatMouse' Trade
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原文地址：http://blog.csdn.net/zp___waj/article/details/46006533
