HDU-1009-FatMouse' Trade(简单贪心!)

FatMouse' Trade

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

Total Submission(s): 57128 Accepted Submission(s): 19145



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

13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

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想起来上回做题，都已经是7月份的事情，正是暑假刚开始那会儿，也不知道怎么了，只能说心意不在了吧。许久未写C++都有些生疏了......

转眼大三，忙着课程，忙着JSP，PHP，忙着做项目，忙着写用例图，当题确乎是没有A了。现在想来，今天也不知怎么，想着做一道题。

就是当时想了很久没做出的这道。居然没用几分钟就解了，算法是一种信念，是程序的灵魂。而A题是提高算法的必经之路。所以，从今

天开始，重新开始吧！Fight!

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
int J,F;
double JF;
}bag[1010];
bool cmp(node x,node y)
{
return x.JF>y.JF;
}
int main()
{
int m,n,i;
while(scanf("%d%d",&m,&n),m!=-1||n!=-1)
{
//The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food
for(i=0;i<n;i++)
{
scanf("%d%d",&bag[i].J,&bag[i].F);
bag[i].JF = bag[i].J*1.0/bag[i].F;
}
//所以这里要是降序排列，因为要用最少的cat food 换取最多的JavaBeans
sort(bag,bag+n,cmp);
double sum = 0;
for(i=0;i<n;i++)
{
if(bag[i].F<m)       //如果FatMouse的M大于当前能换取的值，直接换
{
sum+=bag[i].J;
m-=bag[i].F;
}
else                 //否则还是按照最大的比列能换多少换多少
{
sum+=bag[i].JF*m;
break;
}
}

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