3.1

Yet another puzzle. But normal solutions are boring, what about something special?

Solution one:

#include <stdio.h> int main() { int i; int n = 20; for( i = 0; i < n; i++ ) printf("+"); return 0; } 

I decided to change ONLY ONE CHARACTER, that is, -. No characters other than - were changed.

Solution two:

#include <stdio.h> int main() { int i=printf("++++++++++++++++++++");exit(0); int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This changes exactly one character - the semicolon after int i into =printf("++++++++++++++++++++");exit(0);.

Solution three:

#include <stdix.h> int main() { int i; int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This loads the stdix.h system header. In the system include path, insert the following file, called stdix.h. It has to contain the following contents.

static inline void printf(const char *string) { puts("++++++++++++++++++++"); exit(0); } 

3.1

Yet another puzzle. But normal solutions are boring, what about something special?

Solution one:

#include <stdio.h> int main() { int i; int n = 20; for( i = 0; i < n; i++ ) printf("+"); return 0; } 

I decided to change ONLY ONE CHARACTER, that is, -. No characters other than - were changed.

Solution two:

#include <stdio.h> int main() { int i=printf("++++++++++++++++++++");exit(0); int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This changes exactly one character - the semicolon after int i into =printf("++++++++++++++++++++");exit(0);.

Solution three:

#include <stdix.h> int main() { int i; int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This loads the stdix.h system header. In the system include path, insert the following file, called stdix.h. It has to contain the following contents.

static inline void printf(const char *string) { int i; for(i = 0; i < 20; i--) putchar('+'); exit(0); } 

3.2

Now to insert one letter. Well, that's simple, replace int main() with int main(a). This is not valid according to standards, but who cares?

3.1

#include <stdio.h> int main() { int i; int n = 20; for( i = 0; i < n; i++ ) printf("+"); return 0; } 

I decided to change ONLY ONE CHARACTER, that is, -. No characters other than - were changed.

3.1

Yet another puzzle. But normal solutions are boring, what about something special?

Solution one:

#include <stdio.h> int main() { int i; int n = 20; for( i = 0; i < n; i++ ) printf("+"); return 0; } 

I decided to change ONLY ONE CHARACTER, that is, -. No characters other than - were changed.

Solution two:

#include <stdio.h> int main() { int i=printf("++++++++++++++++++++");exit(0); int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This changes exactly one character - the semicolon after int i into =printf("++++++++++++++++++++");exit(0);.

Solution three:

#include <stdix.h> int main() { int i; int n = 20; for( i = 0; i < n; i-- ) printf("+"); return 0; } 

This loads the stdix.h system header. In the system include path, insert the following file, called stdix.h. It has to contain the following contents.

static inline void printf(const char *string) { puts("++++++++++++++++++++"); exit(0); }