Why this code is showing 4913 a number like ramanujan number?when showing sets are 17 and 2048, 4096 and 17 . Please answer

Question

#include<stdio.h>
int main()
{
int a,count=0,c,j,k,l,m,x;
printf("Enter the number: ");
scanf("%d",&a);
x=a;
for(j=1;j<=a-1;j++){

for(k=1;k<=a-1;k++){
c=(j*j*j)+(k*k*k);
if(c==x)
break;
}
if(c==x){
count++;
break;
}
}
if(count==1){
for(l=1;l<=a-1;l++){
if(l==j||l==k)
continue;
else{

for(m=1;m<=a-1;m++){
c=(l*l*l)+(m*m*m);
if(c==x)
break;
}
}
if(c==x){
count++;
break;}
}

}
if(count==2){
printf("%d is a Ramanujan number. sets are %d and %d, %d and %d.",x,j,k,l,m);
}
else
printf("%d is not a Ramanujan number.",x);
return 0;
}

Answer 1

I would love to try to understand your code, but I swear I can't, so...

There's one solution:

nt Ramanujan(int n) {
int way = 0;

for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
int num = i*i*i + j*j*j;
if (num == n) {
if (!way) printf("%d is a Ramanujan number, sets are: %d and %d", n, i, j);
else printf(", %d and %d", i, j);

way++;
}
if (num > n) break;
}
}

if (!way) printf("%d is not a Ramanujan number.\n", n);
else printf(".\n");

return way;
}

I searched about Ramanujan numbers and there is the 1-way, 2-way, ... and seeing your code I guess you want to find the 2-way Ramanujan numbers, if so, you need to see at the end if the way == 2 and then print it, of course this way you need to keep track of the 4 factors like you were doing.

I hope I helped