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FatMouse‘ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51935 Accepted Submission(s): 17434
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
按照性价比最高的排序，然后求和就行，如果m小于F[i],那么只取这个的a%,而a%=m/F[i].
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct point
{
double x,y;
}a[2000];
bool cmp(point xx,point yy)
{
if((xx.x/xx.y)>(yy.x/yy.y))
return true;
return false;
}
int main()
{
int n;
double m;
while(~scanf("%lf%d",&m,&n))
{
memset(a,0,sizeof(a));
if(m==-1&&n==-1)
break;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
}
sort(a,a+n,cmp);
double sum=0;
for(int i=0;i<n;i++)
{
if(a[i].y<=m)
{
m-=a[i].y;
sum+=a[i].x;
}
else if(m>0)
{
sum+=a[i].x*(m/a[i].y);
m-=a[i].x*(m/a[i].y);
}
}
printf("%.3lf\n",sum);
}
return 0;
}
hdu 1009 贪心
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原文地址：http://blog.csdn.net/qq_24489717/article/details/46482899
