Question

In: Computer Science

Use C language Problem 1: Buy cheese (Level 1) • Problem description You have M dollors...

Use C language

Problem 1: Buy cheese (Level 1)
• Problem description
You have M dollors and want to buy as much cheese as possible. There is a cheese shop with
two kinds of cheese. The unit prices of the both cheese are p1 and p2, the total amount of them
are a1 and a2. You can buy either cheese or both, but the amount you buy cannot exceed its total
amount. Write a program to output the largest amount of cheese you can buy.
• Input & output requirements
All the integers variables (M, p1, a1, p2, and a2) are taken from user input. The input begins
with a line containing one non-negative interger M, and followed by two lines, each line
contains two integers, unit price and total amount.
The output should be a real number accurate up to 2 decimal places.
• Sample results
Sample 1
Please input M: 2↲
Please input p1 and a1: 1 2↲
Please input p2 and a2: 2 2↲
The maximum amount of cheese is 2.00
Sample 2
Please input M: 3↲
Please input p1 and a1: 2 2↲
Please input p2 and a2: 1 1↲
The maximum amount of cheese is 2.00

Solutions

Expert Solution

Note:AS PER THE QUESTION THE ANSWER CAN BE FRACTIONAL/FLOAT ALSO.

EXAMPLE;

4
1 3
2 2

ANSWER IS 3.50.(3 items from 1st type and 0.5 items from second)

#include <stdio.h>

void main()
{
int a1,a2,p1,p2;
float dollars,ans;
scanf("%f",&dollars);
scanf("%d %d",&p1,&a1);
scanf("%d %d",&p2,&a2);
if(p1<p2) //will consider lowest unit price first
{
  
while(a1 && dollars>=p1)
{
dollars=dollars-p1;
ans=ans+1;a1--;
}
if(a1>0) //still items left
{   
printf("%.2f",ans+dollars/p1);
return;
}
while(a2 && dollars>=p2) //if first type exhausted,will go for next
{
dollars=dollars-p2;
ans=ans+1;a2--;
}
if(a2>0)
{
printf("%.2f",ans+dollars/p2);
return;
}
}

else //same process but p1>=p2;either above or this part!!!
{
while(a2 && dollars>=p2)
{
dollars=dollars-p2;
ans=ans+1;a2--;
}
if(a2>0)
{
printf("%.2f",ans+dollars/p2);printf("secondYes");
return;
}
while(a1 && dollars>=p1)
{
dollars=dollars-p1;printf("Yes");
ans=ans+1;a1--;
}
if(a1>0)
{   
printf("%.2f",ans+dollars/p1);
return;
}
}

printf("%.2f",ans);
return;
}


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