C program to represent numbers in numerator and denominator in string format-C++-php.cn

C program to represent numbers in numerator and denominator in string format

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Release： 2023-09-01 16:41:07

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C program to represent numbers in numerator and denominator in string format

Question

Find the fractional part from two integers given by the user at runtime and represent it in string format by using dynamic memory allocation Numerator and denominator.

Solution

The solution to express the numerator and denominator in string format is as follows:

Example-

The input is as follows:

Numerator1 = 3
Denominator2 = 2

numerator2 = 4
denominator2 = 7

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The output result is as follows −

Fractional part1: 1.5
Fractional part2: 0.(571428)

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Example

The following is a program written in C for Representing the numerator and denominator in string format −

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
char* fractionToDecimal(int numerator, int denominator) {
   char *p;
   int psz, n, *dec, dsz, x;
   long long num, den, k, f;
   int i, repeat_at;
   int neg = 0;
   psz = dsz = 100; n = x = 0;
   p = malloc(psz * sizeof(char));
   //assert(p);
   neg = ((numerator > 0 && denominator < 0) ||
   (numerator < 0 && denominator > 0)) ? 1 : 0;
   num = numerator;
   den = denominator;
   num = (num < 0) ? -num : num;
   den = (den < 0) ? -den : den;
   k = num / den;
   f = num % den;
   if (neg && (k || f)) p[n ++] = &#39;-&#39;;
      n += sprintf(&p[n], "%lld", k);
   if (!f) {
      p[n] = 0;
      return p;
   }
   p[n ++] = &#39;.&#39;;
   dec = malloc(dsz * sizeof(int));
   repeat_at = -1;
   if (f < 0) f = -f;
   while (f) {
      for (i = 0; i < x; i += 2) {
         if (dec[i] == f) {
            repeat_at = i;
            goto done;
         }
      }
      if (x + 1 >= dsz) {
         dsz *= 2;
         dec = realloc(dec, dsz * sizeof(int));
      }
      dec[x ++] = f;
      f *= 10;
      k = f / den;
      dec[x ++] = k;
      f = f % den;
   }
   done:
   for (i = 0; i < x; i += 2) {
      if (n + 3 > psz) {
         psz *= 2;
         p = realloc(p, psz * sizeof(char));
      }
      if (repeat_at == i) {
         p[n ++] = &#39;(&#39;;
      }
      p[n ++] = &#39;0&#39; + dec[i + 1];
   }
   if (repeat_at != -1) p[n ++] = &#39;)&#39;;
      p[n ++] = 0;
   free(dec);
   return p;
}
int main(void){
   int n,d;
   printf("enter numerator1 and denominator1:");
   scanf("%d%d",&n,&d);
   printf("n = %d, d = %d ", n, d);
   printf("</p><p>Fractional part1: %s </p><p>",fractionToDecimal(n, d));
   printf("enter numerator2 and denominator2:");
   scanf("%d%d",&n,&d);
   printf("</p><p>n = %d, d = %d ", n, d);
   printf("</p><p>Fractional part2: %s</p><p> ",fractionToDecimal(n, d));
   return 0;
}

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Output

When the above program is executed, it produces the following results−

enter numerator1 and denominator1:4 5
n = 4, d = 5
Fractional part1: 0.8
enter numerator2 and denominator2:5 9

n = 5, d = 9
Fractional part2: 0.(5)

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