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 很简单的贪心题， 先按照价值/代价的比值来排序，肯定是先买比值大的。
#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; }
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