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sn#540145 filedate 1980-10-08 generic text, type C, neo UTF8
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C00002 00002 Exercise
C00004 00003 Sample Program
C00007 00004 Sample Program
C00010 ENDMK
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Exercise
Write a program, using indefinite iterations (WHILE ... DO) only,
that reads a sequence of positive integers terminated by a zero sentinel.
For each number N, the program should find the smallest non-negative
integers X and Y for which Xā2 - Yā2 = N, or announce that there are
no such numbers. Try the program on
21 9 180 4641 32 101 0
Watch out for non-terminating iterations (``infinite loops''). If you have
trouble, do these problems as warmups:
(1) Read and print the input data, as above, without printing the sentinel.
(2) Read one positive number N and find X, Y as above.
Remember, whenever you write an indefinite iteration, to find a convincing reason
to believe it will terminate. If it isn't obvious, put it in a comment in the
program.
Sample Program
EXAMPLE OF THE USE OF INDEFINITE ITERATION, AND INPUT FROM
TERMINAL. THIS PROGRAM COMPUTES A SQUARE ROOT WITHOUT
USING THE STANDARD FUNCTION `SQRT' .
PROGRAM DEMOWHILE ;
VAR X, (* VALUE OF WHICH SQUARE ROOT IS TO
BE TAKEN *)
ROOT : REAL ; (* SQUARE ROOT OF X *)
BEGIN (* HEADER AND INPUT *)
WRITELN ( TTY, ' THIS PROGRAM COMPUTES SQUARE ROOTS' ) ;
WRITELN ( TTY, ' WITHOUT USING THE BUILT-IN FUNCTION' ) ;
WRITE ( TTY, ' INPUT NUMBER : ' ) ;
BREAK ( TTY ) ;
READ ( TTY, X ) ;
(* INPUT VALIDATION *)
WHILE X < 0.0 DO
BEGIN
WRITELN ( TTY, ' NEGATIVE NUMBERS NOT ALLOWED ' ) ;
WRITE ( TTY, ' TRY AGAIN : ' ) ;
BREAK ( TTY ) ;
READ ( TTY, X )
END ;
(* INITIALIZATION *)
ROOT := 1.0 ;
(* COMPUTATION OF SQUARE ROOT *)
WHILE ABS( (ROOT - X/ROOT) ) > 1.0E-6 DO
ROOT := 0.5 * ( ROOT + X/ROOT ) ;
(* OUTPUT *)
WRITE ( TTY, ' SQUARE ROOT OF ', X :8 :4, ' IS ', ROOT :10 :6 )
END.
THIS PROGRAM COMPUTES SQUARE ROOTS
WITHOUT USING THE BUILT-IN FUNCTION
INPUT NUMBER : -2.0
NEGATIVE NUMBERS NOT ALLOWED
TRY AGAIN : 2.0
SQUARE ROOT OF 2.0000 IS 1.414213
Sample Program
READING IN A NUMBER IN BASE 7 AND CONVERTING IT TO BASE 10
THIS PROGRAM HAS SOME SUBTLE POINTS TO IT:
1. THE PARENTHESES AROUND " ORD(C) - ORD(0) " TO
PREVENT INTEGER OVERFLOW.
2. WHEN YOU TRY TO READ A CHARACTER AT THE END OF THE
END OF A LINE, YOU GET A SPACE. THIS IS THE REASON
WHY THE PROGRAM ALWAYS STOPS AT THE END OF A LINE.
PROGRAM CONVERT ( INPUT*, OUTPUT ) ;
VAR C : CHAR ; (* CHARACTER LAST READ IN. *)
VALUE : INTEGER ; (* VALUE IN BASE 10. *)
BEGIN
C := ' ' ;
WHILE ( C< '0' ) OR ( C> '6' ) DO READ ( C ) ;
VALUE := ORD ( C ) - ORD ( '0' ) ;
READ ( C ) ;
WHILE ( C >= '0' ) AND ( C <= '6' ) DO
BEGIN
VALUE := VALUE*7 + ( ORD (C) - ORD ('0') ) ;
READ ( C )
END;
WRITELN ( ' DECIMAL VALUE = ', VALUE :8 )
END.
INPUT: 46
DECIMAL VALUE = 34
INPUT: ///// 23 //////
DECIMAL VALUE = 17