Numbers in base 10 that are palindromic in bases 2, 4, and 16: Difference between revisions
Numbers in base 10 that are palindromic in bases 2, 4, and 16 (view source)
Revision as of 22:25, 24 June 2021
, 2 years agoAdd PL/M
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23 found: 0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365 3855 4095 4097 4369 12291 13107 20485 21845
</pre>
=={{header|PL/M}}==
<lang plm>100H:
/* CP/M CALLS */
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
DECLARE MAXIMUM LITERALLY '25$000';
/* PRINT A NUMBER */
PRINT$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... $');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT$NUMBER;
/* REVERSE NUMBER GIVEN BASE */
REVERSE: PROCEDURE (N, B) ADDRESS;
DECLARE (N, R) ADDRESS, B BYTE;
R = 0;
DO WHILE N > 0;
R = R*B + N MOD B;
N = N/B;
END;
RETURN R;
END REVERSE;
/* CHECK IF NUMBER IS PALINDROME */
PALIN: PROCEDURE (N, B) BYTE;
DECLARE N ADDRESS, B BYTE;
RETURN N = REVERSE(N, B);
END PALIN;
DECLARE I ADDRESS, C BYTE;
C = 0;
DO I = 0 TO MAXIMUM;
IF PALIN(I,2) AND PALIN(I,4) AND PALIN(I,16) THEN DO;
CALL PRINT$NUMBER(I);
C = C + 1;
IF C = 15 THEN DO;
CALL PRINT(.(13,10,'$'));
C = 0;
END;
END;
END;
CALL EXIT;
EOF</lang>
{{out}}
<pre>0 1 3 5 15 17 51 85 255 257 273 771 819 1285 1365
3855 4095 4097 4369 12291 13107 20485 21845</pre>
=={{header|Python}}==
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