programming puzzlesWrite a "Hello World" program in 'C' without using a semicolon.
(this is tricky)
C/C++ : Exchange two numbers without using a temporary variable.
(somehow classic way)
C/C++ : Find if the given number is a power of 2.
(somehow classic way)
Write a program whose printed output is an exact copy of the source. Needless to say, merely echoing the actual source file is not allowed.
(a bit hard to think)
source: http://developers.slashdot.org/article.pl?sid=04/12/04/0116231[吴永铮 (4-24 17:12, Long long ago)]
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interesting... let me try 1&2...1.
#include <stdio.h>
void main() {
if (printf("hello world\n")) {}
}
2.
a=a+b;
b=a-b;
a=a-b;
[hash (4-24 18:10, Long long ago)]
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(引用 hash:interesting... let me try 1&2...1. #include void main() { if (printf("hello world\n")) {} } 2. a=a+b; b=a-b; a=a-b; )世界真奇妙[soc胖浪人 (4-25 0:27, Long long ago)] [ 传统版 | sForum ][登录后回复]3楼
what do you mean by classic way?C/C++ : Find if the given number is a power of 2.
(somehow classic way)
can i check bit by bit?[icky (4-25 11:33, Long long ago)]
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(引用 hash:interesting... let me try 1&2...1. #include void main() { if (printf("hello world\n")) {} } 2. a=a+b; b=a-b; a=a-b; )oh, i was thinking using whilei did not know a pair of empty brackets can follow "if"[icky (4-25 11:34, Long long ago)] [ 传统版 | sForum ][登录后回复]5楼
(引用 icky:what do you mean by classic way?C/C++ : Find if the given number is a power of 2. (somehow classic way) can i check bit by bit?)i was thinking of checking bit also.no. 4 is tricky...[hash (4-25 11:36, Long long ago)] [ 传统版 | sForum ][登录后回复]6楼
(引用 icky:what do you mean by classic way?C/C++ : Find if the given number is a power of 2. (somehow classic way) can i check bit by bit?)(3) can be done by less than 5 instructionsBy classic (or classical?), I mean a lot of people use the method.[吴永铮 (4-25 12:26, Long long ago)] [ 传统版 | sForum ][登录后回复]7楼
(引用 吴永铮:(3) can be done by less than 5 instructionsBy classic (or classical?), I mean a lot of people use the method.)check equality with all 2's powerthere are at most 32 of them...[icky (4-25 12:56, Long long ago)] [ 传统版 | sForum ][登录后回复]8楼
(引用 icky:check equality with all 2's powerthere are at most 32 of them...)can be simpler and faster[吴永铮 (4-25 15:26, Long long ago)] [ 传统版 | sForum ][登录后回复]9楼
(引用 吴永铮:can be simpler and faster)先模1<<16和除 1<<16两边应该有一边是0，如果两边都不是0，一定就不是2的power了
然后把不为0的那一边再模1<<8和除1<<8
当然这里假设是unsigned int, 32bits
[icky (4-25 16:46, Long long ago)]
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(引用 icky:先模1...)(num > 0) && (((num - 1) & num) == 0) ?[这就是生活 (4-26 0:31, Long long ago)] [ 传统版 | sForum ][登录后回复]11楼
(引用 这就是生活:(num > 0) && (((num - 1) & num) == 0) ?)correctjust ((num - 1) & num) is enough
((num - 1) ^ num) also can
It's quite practical and quite frequent. It's important to know it, otherwise you don't understand what's going on.[吴永铮 (4-26 0:50, Long long ago)]
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(引用 吴永铮:correctjust ((num - 1) & num) is enough ((num - 1) ^ num) also can It's quite practical and quite frequent. It's important to ...)opps ^ is wrong. It's always true.[吴永铮 (4-26 1:06, Long long ago)] [ 传统版 | sForum ][登录后回复]13楼
my answer to (4)#include <stdio.h>
int main (void)
{
const char *str = "#include <stdio.h>%cint main (void)%c{%c const char *str = %c%s%c;%c printf(str, 10, 10, 10, 34, str, 34, 10, 10, 10, 10);%c return 0;%c}%c";
printf(str, 10, 10, 10, 34, str, 34, 10, 10, 10, 10);
return 0;
}[吴永铮 (4-27 21:56, Long long ago)]
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(引用 吴永铮:my answer to (4)#include int main (void) { const char *str = "#include %cint main (void)%c{%c const char *str = %c%s%c;%...)(Y)[hash (4-29 4:42, Long long ago)] [ 传统版 | sForum ][登录后回复]15楼