Thread: I need help for my homework. [so important]

As Salem said... break your code in small portions. Here's my implementation with comments for your study:

Code:

// cobalt.c
//
// Compile and link:
//
//  $ gcc -O2 -o cobalt cobalt.c  # add -lm if you use exp().
//

#include <stdio.h>
#include <stdlib.h>   // ony for EXIT_* symbols.

// FIX: Don't need this with precalculated constant.
//#include <math.h>

// Instead of exp(-0.693)/5.272, this C is precalculated printing
// this expression with 60 decimal places in double precision:
//
//  double C = exp(-0.693)/5.272;
//  printf( "C=%.60f\n", C );
//
#define C 0.09485462740815016335904630295772221870720386505126953125 

int main ( void )
{
  int year;
  double initial;

  printf ( "Initial amount of Cobalt (grams)> " );
  fflush( stdout ); // previous print doesn't have a final '\n',
                    // so, it's prudent to flush stdout.

  // Check for your inputs!
  if ( scanf ( "%lf", &initial ) != 1 )
  {
error:
    fprintf( stderr, "Wrong amount.\n" );
    return EXIT_FAILURE;
  }

  if ( initial <= 0.0 )
    goto error;

  printf ( " Years     Amount      Decay\n" );
  //       "yyyyyy  aaaaaa.aa  dddddd.dd\n"
  //         "% 6d  % 9.2f  % 9.2f"

  // Calculate and show decay year by year.
  for ( year = 0; year <= 5; year++ )
  {
    double remain, decay;

    decay = C * year;
    remain = initial * ( 1.0 - decay );

    // We can format the output directly on the format argument.
    // no need to use ' ' as arguments! See comment above, after printing
    // the 'header'.
    printf ( "% 6d  % 9.2f  % 9.2f\n", year, remain, decay );
  }

  return EXIT_SUCCESS;
}

[]s
Fred