plot - Print a representation of the Pythagorean Triple in C -

i trying create program prints mapping data found pythagorean triples in c. far, have coded program able find triples.

#include <stdio.h> #include <math.h>  int main (int argc, char * argv[]) {     int a, b, c;     int a2, b2, c2;     int limit = 60;      (a = 1; <= limit; a++) {         a2 = * a;         (b = 1; b <= limit; b++) {             b2 = b * b;             (c = 0; c <= ((int)sqrt(a2+b2)); c++) {                 c2 = c * c;                 if (a < b && (c2 == (a2 + b2))) {                     printf("triple: %d %d %d\n", a, b, c);                 }             }         }     } } 

the intended output expected in format:

123456789012345678901234567890123456789012345678901234567890 1\ 2 \ 3  \ 4  *\ 5    \ 6     \ 7      \ 8     * \ 9        \ 0         \ 1          \ 2    *   *  \ 

i trying write loop this, cannot think of how print in way. suggestions?

update: managed print x , y axis (x = , y = b). values correct, mapping part left.

    (int x = 0; x < a; x++) { // x-axis =         printf("%d ", x);     }      printf("\n");      (int y = 0; y < b; y++) { // y-axis = b          printf("%d\n", y);     } 

update: modified code, output printing, having problems printing spaces. tried manually adding spaces " \ " , " * " distorts whole image.

#include <stdio.h> #include <math.h> #include <stdbool.h>  bool is_perfect_square(int num);  int main (int argc, char * argv[]) {      int a, b, c;     int a2, b2, c2;     int limit = 60;     bool flag = false;      (a = 1; <= limit; a++) {         a2 = * a;         (b = 1; b <= limit; b++) {             b2 = b * b;             (c = 0; c <= ((int)sqrt(a2+b2)); c++) {                 c2 = c * c;                 if (a < b && (c2 == (a2 + b2))) {                     // printf("triple: %d %d %d\n", a, b, c);                 }             }         }     }      (int x = 0; x < a; x++) {         (int y = 0; y < b; y++) {             if (x == 0) {                 printf("%d ", ((y+1)% 10));             } else if (y == 0) {                 printf("%d ", (x % 10));             } else if (x == y) {                 printf("\\");             } else if (is_perfect_square((x*x) + (y*y))) {                 printf("*");             }         }         printf("\n");     } }  bool is_perfect_square (int num) {     int root = (int)(sqrt(num));      if (num == root * root) {         return true;     } else {         return false;     } } 

still working on possible solution.

hint :

have nested loop index i,j;

for 0.. limit {   j 0 .. limit {    /*implement below algorithm here!!*/   }     printf("\n") } 

algorithm used inside loop:

• if i == 0 print x axis values printing (j+1)% 10 [see note @ end]
• else if j == 0 print y axis values printing i % 10
• else if i == j print '\'
• else if is_perfect_square((i*i) + (j*j)) returns 1, print '*'
• else print space.

specifications is_perfect_square function: function returns 1 if input perfect square , 0 otherwise. example:

• is_perfect_square(25) should return 1
• is_perfect_square(7) should return 0
• is_perfect_square(49) should return 1

note: i == 0 case should print j%10 start output 0 represent origin. output provided in question starts 1. hence using (j+1)%10

you might need handle corner cases, should straight forward once algorithm implemented in code.