FatMouse' Trade
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 20 Accepted Submission(s) : 14
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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
13.333
31.500
Author
CHEN, Yue
Source
很簡單的貪心題,
先按照價值/代價的比值來排序,肯定是先買比值大的。
#include
#include
using namespace std;
struct stu
{
int a;
int b;
double rate;
}s[1005];
bool cmp(stu x,stu y)
{
if(x.rate!=y.rate)
return x.rate>y.rate;
else
return x.a=s[i].b)
{
m-=s[i].b;
cnt+=s[i].a;
}
else
{
cnt+=s[i].rate*m;break;
}
}
printf("%.3lf\n",cnt);
}
return 0;
}