HDU--1009 -- FatMouse' Trade

 

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 33514    Accepted Submission(s): 10902

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.

 
[align=left]Output[/align]
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

 

 

Code:

#include"cstdio"
#include"cstdlib"
#include"algorithm"
#include"iostream"
using namespace std;

typedef struct changeinfo
{
int j;
int f;
}changeinfo;    //房间信息的结构体

bool cmp(changeinfo a,changeinfo b)
{
return 1.0*a.j/a.f > 1.0*b.j/b.f;//乘以1.0使之变为double型
}

int main()
{
int m,n,i,k;
double sum;
changeinfo a[1001];
while(scanf("%d%d",&m,&n),(m+1)&&(n+1))
{
sum=0;

for(i=0;i<n;i++)
scanf("%d%d",&a[i].j,&a[i].f);

sort(a,a+n,cmp);    //根据相除结果从大到小排序,参数依次为参与排序的首尾地址和cmp比较函数

//for(i=0;i<n;i++)
//    printf("%d %d\n",a[i].j,a[i].f);

if(m==0)//==!题目的问题需要考虑这两种情况才能AC
{
for(i=0;i<n;i++)
if(a[i].f==0)
sum = sum + a[i].j;
}
if(n==0)
sum = 0;
if(m!=0 && n!=0)
{
for(i=0,k=m;i<n,k>=0;i++)
{
if(a[i].f>k)        //剩余猫粮不足的情况
sum = (1.0*a[i].j/a[i].f)*k + sum;
else
sum  = a[i].j + sum;
k = k - a[i].f;
}
}

printf("%.3lf\n",sum);

}
return 0;
}