0
Mentor
Can you spot a number that divides both 34 and 24? Assuming the answer is yes, what would you then do? Would you do it again, potentially? Would you want to do that a fixed number of times regardless of the numbers, or would you repeat the operation until a defined end condition was reached?
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Mentor
Does posting a duplicate thread mean that you can't be arsed working this through and just want someone to give you the answer?
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Contributor
Here.....I have the solution of your first assignment....

Code:
```#include<stdio.h>
#include<conio.h>

int gcd(int a, int b)
{
int c,d,m,g=1;
for(m=1;m<=(a<b?a:b);m++)
{
c=a%m;
d=b%m;
if((c==0)&&(d==0)&&(m>g))
{
g=m;
}
}
return(g);
}

void main()
{
int a,b,c,d,p,q,r,s;
printf("Enter a,b,c,d");
scanf("%d%d%d%d", &a, &b, &c, &d);
printf("\nFirst fraction %d/%d\nSecond fraction %d/%d\n", a,b,c,d);
q=(b*d)/gcd(b,d);
p=a*(q/b) + c*(q/d);
r=p/gcd(p,q);
s=q/gcd(p,q);
printf("\nThe fraction output is=%d/%d\n", r,s);
getch();
}```
---------------
@ r k @
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Contributor
sorry.....I thought the first one was not posted properly....I am extremely sorry....
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Mentor
Oh well, what the heck. Here's what I got for the second assignment. It was fun to solve:
Code:
```void go4e_15046b()
{
/* ~~~
1/3 = .(3)             > 0.(3)
22/5 = 4.4             > 4.4
1/7 = .(142857)        > 0.(142857)
3/8 = .375             > 0.375
45/56 = .803(571428)   > 0.803(571428)

Test your program with the fractions above and the fraction 11/59.
0.(1864406779661016949152542372881355932203389830508474576271)
_0.18644067796610169491525423728814 <- Windows Calc
~~~
*/
int N=11,D=59;
printf("Enter N: %d\n",N);
//char buf[32];
//fgets(buf,30,stdin);
//N=atoi(buf);
printf("Enter D: %d\n",D);
//fgets(buf,30,stdin);
//D=atoi(buf);

printf("\nN=%d, D=%d\n\n",N,D);
int A=0;
if (N>D)
{
A=N/D;
N-=A*D;
}
const int COUNT=100;
int loop[COUNT+5];
char result[COUNT+5]; // +5 to avoid any buffer overflows if we hit COUNT
int resi=0;
int next=0,loopf=-1;
// Long division algorithm
for (int max=COUNT; max && loopf<0; max--)
{
N*=10;
for (int i=0; i<next; i++)
{
if (N==loop[i])
{
loopf=i;
break;
}
}
// we don't want to concatenate N if we've found a loop
if (loopf<0)
{
loop[next++]=N;
int H=N/D; // how many times does D go into N?
result[resi++]=H+'0';
result[resi]=0;
N=N-H*D;
if (!N) break;
}
}
printf("result before the brackets: '%s'\n",result);
if (loopf>=0)
{
int i;
strcat(result,")");
// find the end of the result
for (i=0; result[i]; i++) ;
for (; i>=loopf; i--)
{
result[i+1]=result[i];
}
result[i+1]='(';
}
printf("%d.%s\n",A,result);
}```
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Contributor
Brilliant job....xpi0t0s!!! Congrats....It was really fun solving 'em...