C语言编程题 FatMouse' TradeFatMouse' TradeTime Limit:1000MS Memory Limit:65536KTotal Submit:395 Accepted:95 Description FatMouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his favorite food,JavaB

C语言编程题 FatMouse' TradeFatMouse' TradeTime Limit:1000MS Memory Limit:65536KTotal Submit:395 Accepted:95 Description FatMouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his favorite food,JavaB
C语言编程题 FatMouse' Trade
FatMouse' Trade
Time Limit:1000MS Memory Limit:65536K
Total Submit:395 Accepted:95
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
下面是我的代码 我试了多组程序,但OJ上老是通不过.
#include
void main()
{int m,n,i,j;
float a[1000][2],b[1000],t,s,w;
while(scanf("%d%d",&m,&n),=-1||n!=-1)
{s=0;w=0;
for(i=0;i

C语言编程题 FatMouse' TradeFatMouse' TradeTime Limit:1000MS Memory Limit:65536KTotal Submit:395 Accepted:95 Description FatMouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his favorite food,JavaB
//题目的意思应该是将所有的case统一输出而不是计算一个输出一个
#include
void main()
{
int m,n,i,j,s=0;
float a[1000][2],b[1000],t,w;
float res[1000];
while(scanf("%d%d",&m,&n),m!=-1||n!=-1)
{
w=0;res[s]=0;
for(i=0;i