Help me with C programs

Re: help me with my c programmes

Hi, How about showing us your efforts?

i can't solve those two programmes so that i post them here 😁

abbd1990

i can't solve those two programmes so that i post them here 😁

ok, not a problem. C users will post them for you here 😉. By the way will Java programs work for you?

What biggie meant was, can you try typing the basic C structure first (eg the "#include" preprocessor directive)? Then we can see what area you are struggling at 😀 You'll learn better if you show us your attempts.

Unless, you have no clue at all?

I only wanted him to try. Once you know the formula for generating perfect numbers; rest is cakewalk.

I'd still want him to post the incorrect program and ask CEans to guide him than looking for a ready-made program.

The_Big_K

I only wanted him to try. Once you know the formula for generating perfect numbers; rest is cakewalk.

I'd still want him to post the incorrect program and ask CEans to guide him than looking for a ready-made program.

Even I also don't wanted to give him/her ready made program but I don't even want to give him/her a bad impression from this forum in the starting itself. I want him/her to be more active.😀

Hey abbd1990, check these two small programs. Are these two what you wanted? 😀

Write a program that outputs all perfect numbers less than 1000

/*
C program to display all perfect numbers less than 1000
*/
#include <stdio.h>
#include <conio.h>
 
int main() {
 
              int i,j,num,sum;
              clrscr(); // To clear the screen
 
              printf("The perfect numbers less than 1000 are:");
              for(i=1;i<1000;i++) {
                     num=i;
                     sum=0;
 
                     for(j=1;j<num;j++) {
                               if(num%j==0)
                                      sum=sum+j;
                               }
                     }
 
                     if(sum==num)
                               printf("\n%d",num);
             }
             getch(); 
             return 0;
}

Write a program that inputs two fractions in the form a/b and c/d , and outputs their sum in the form p/q cancelled down to its simplest form ?

/*
C program to accept two fractions in the form a/b and c/d and display 
their sum in the form p/q cancelled down to its simplest form
*/
 
#include <stdio.h>
#include <conio.h>
 
int main() {
 
              int nr,nr1,nr2,dr,dr1,dr2,i;
              char nrsign=' ',drsign=' ';
 
              clrscr(); //To clear the screen
 
              printf("\nFirst fraction :");
              printf("\n\tInput Numerator:");
              scanf("%d",&nr1);
              printf("\n\tInput Denominator:");
              scanf("%d",&dr1);
              printf("\nSecond fraction :");
              printf("\n\tInput Numerator:");
              scanf("%d",&nr2); 
              printf("\n\tInput Denominator:");
              scanf("%d",&dr2);
              printf("\nYour Inputs:");
              printf("\n\tFirst fraction : %d/%d",nr1,dr1);
              printf("\n\tSecond fraction : %d/%d",nr2,dr2);
 
              nr=nr1*dr2+nr2*dr1;
              dr=dr1*dr2;
 
              if(nr<0) {
                          nrsign='-';
                          nr=-nr;
              }
              if(dr<0) { 
                          drsign='-';
                          dr=-dr;
              }
 
              i=2;
              while(i<=dr){
 
                          if(nr%i==0 && dr%i==0){
                                  nr=nr/i;
                                  dr=dr/i;
                                  i=2;
                          }
                          else i=i+1;
              }
 
              printf("\nResult : %c%d /%c%d",nrsign,nr,drsign,dr);
              getch();
              return 0;
}

Hey buddy you got the code but if you would have tried and posted incorrect code we all would have been much more happy anyways welcome to the forum and we hope better efforts from those joining in but if you have any problem we are always there to help you

thanks very much my friends for your kind help