作业帮帮我看看这为什么在oj上就是不行 FatMouse' TradeDescriptionFatMouse 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
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帮我看看这为什么在oj上就是不行 FatMouse' TradeDescriptionFatMouse 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
帮我看看这为什么在oj上就是不行 FatMouse' Trade
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
大神应该懂的
下面是我的代码
#include
struct stu
{
double a[2];
double b;
}stu[10000],t;
int main()
{
int i,j,n;
double m;
double bb=0;
while(scanf("%lf%d",&m,&n)&&(m!=-1||n!=-1))
{
for(i=0;i
帮我看看这为什么在oj上就是不行 FatMouse' TradeDescriptionFatMouse 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
for(i=0;i