YX Hangzhou Electric oj 1008

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33442    Accepted Submission(s): 10862

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

AC codes + detailed explanation:

    
    
# The include <iostream> // greedy
# The include <cstdio> // title meaning: FatMouse mice exchange their favorite food with a m g cat food and cat food like mice but in many warehouses and warehouses of each exchange rate is not the same, but the exchange rate for each J [i] g cat against F [i] food (i is the number of layers)
#include<cstring>
const int MAX= 10001;
int J[MAX];
int F[MAX];
double p[MAX];
using namespace std;
int main()
{
int n,m,i,j;
double sum;
while( cin>>m>>n&&(m!= -1&&n!= -1))
{
for(i= 1;i<=n;i++)
{
cin>>J[i]>>F[i];
P [I] = Double (J [I]) / F. [I]; // each exchange rates
}
sum= 0;
the while (m! = 0 && n-! = 0 ) // 0 or food is out of the loop when the cat 0
{
int k;
double Max= -1;
for (I = . 1 ; I <= n-; I ++) // each selected to be a rate of the maximum number of layers (room)
{
if(p[i]>Max)
{
Max=p[i];
k=i;
}
}
IF (mF of [K]> = 0 ) // if there is remaining cat food or just
{
= F. M- [K]; // needed subtracting cat
+ J = SUM [K]; // add to the food exchange
P [K] = 0 ; // layer against the exchange rate that is over 0 - no
}
the else // If the cat is not enough, use the rest of the cat food all to exchange food
{
+ P = SUM [K] * m; // All food exchange
= m 0 ; // make the cat food is 0, the redemption is completed - out of the loop
}
}
the printf ( "% .3lf \ n-" , SUM); // Output
}
return 0;
}
    
    

 

Reprinted from (https://blog.csdn.net/ecjtu_yuweiwei/article/details/9427885)