Thread: math problem

hi there im trying to make a math program
the input is

a/b # c/d

# = / or + or - or *


example

input
1/4 / 1/4
output
1
input
2/7 - 4/14
output
0
input
4/5 + 4/7
output
1 13/35
input
2/3 * 4/5
output
8/15


this is my code for only + sign math program

Code:

#include <stdio.h>
#include <conio.h>

int main(void){
    int a, b, c, d, e, f, g;
    
    scanf("%d/%d%d/%d", &a, &b, &c, &d);
    b > 0;
    d > 0;
    
    e = (a * d) + (b * c);
    f = b * d;
    
    if(e < f)
    printf("%d/%d + %d/%d = %d/%d", a, b, c, d, e, f);
    if(e > f)
    printf("%d/%d + %d/%d = %d %d/%d", a, b, c, d, e / f, e % f, f);
    if(e == f)
    printf("%d/%d + %d/%d = 1", a, b, c, d);
    
    return 1;
    }

my question is how to replace # in input with / or * or + or -

sorry for my bad english tq

hi there im trying to make a math program
the input is

a/b # c/d

# = / or + or - or *

example

input
1/4 / 1/4 = 0 / 0 = ?
output
1

input
2/7 - 4/14 = 0 - 0 = 0
output
0

input
4/5 + 4/7 = 0 + 0 = 0
output
1 13/35

input
2/3 * 4/5 = 0 * 0 = 0
output
8/15

you have declared a,b,c,d,e as integer variables which gives output as 0 for these situations.

had it been (5/4)/(8/3) = (1.25) / (2.666) = (1) / (2) = 0

corrected code is

Code:

#include <stdio.h>

int main(void)
{
    int a, b, c, d, e;
	char hash;
    
 	printf("\n\nEnter a,b,c,d : ");   
	scanf("%d%d%d%d", &a, &b, &c, &d);
    
	printf("Enter # : ");	
	
	fflush(stdin);
	scanf("%c",&hash);
	
	switch(hash)
	{
		case '/' :
				e = (a/b)/(c/d);
				break;	
		case '*' :
				e = (a/b)*(c/d);				
				break;
		case '+' :
				e = (a/b)+(c/d);				
				break;	
		case '-' :
				e = (a/b)-(c/d);				
				break;
		default :
				printf("\n\nInvalid operator");
	
	}
	
	printf("\n\n   (%d/%d)%c(%d/%d) = %d \n\n",a,b,hash,c,d,e);

    return 0; 
}

thank you

thx man!
now i understand

Don't ever fflush(stdin)

why ?