Including: Fill & Download for Free

And here's some bonus solutions!#include <stdlib.h>  int main() {  /* Cross your fingers and hope seq exists! */  return system("seq 1 100"); } The below one works on my system, can't guarantee results though.#include <iostream> #include <stdlib.h>  int num; void(**rptr)();  void foo() {  if(num >= 100) exit(0);  std::cout << ++num << std::endl;  *rptr++ = foo; }  int main() {  rptr = (void(**)())alloca(sizeof(*rptr) * 200) - 1;  foo();  return 0; } And if order is not your thing you might try#include <iostream> #include <stdlib.h>  int main() {  int x = 0;  x |= !fork() << 0;  x |= !fork() << 1;  x |= !fork() << 2;  x |= !fork() << 3;  x |= !fork() << 4;  x |= !fork() << 5;  x |= !fork() << 6;  if(1 <= x && x <= 100) std::cout << x << std::endl;  return 0; } And similar to bashing templates you can bash the preprocessor#include <stdio.h>  #define F4 "%d\n%d\n%d\n%d\n" #define F20 F4 F4 F4 F4 F4 #define F100 F20 F20 F20 F20 F20  #define X4(y) , y, y + 1, y + 2, y + 3 #define X20(y) X4(y) X4(y + 4) X4(y + 8) X4(y + 12) X4(y + 16) #define X100(y) X20(y) X20(y + 20) X20(y + 40) X20(y + 60) X20(y + 80)  int main() {  printf(F100 X100(1));  return 0; } New (July 22nd 2014) - Advanced preprocessor bashing#if (__COUNTER__ == 0) #include <stdio.h> int main() {  #include __FILE__  return 0; } #elif (__COUNTER__ < 300) printf("%d\n", __COUNTER__ / 3); #include __FILE__ #endif Anyone up for writing the book '101 ways to write 1 to 101: C edition'?

Short answer: No.{0} is an element of {{0}}, and it’s an element of {{0}, 0}, but it is not an element of {0}.The elements of {0} includes only the value 0.

It depends on your domain.For any value [math]x, (100 - x) + x = 100[/math]. So the pair [math](x, y)[/math] where [math]y = (100-x)[/math] is a pair with sum [math]100[/math].Now, the question is, what is the domain of [math]x[/math] and [math]y[/math]? That is, what values can [math]x[/math] and [math]y[/math] take in your assumption?If you limit [math]x[/math] and [math]y[/math] to be natural numbers (0 included), then you have 51 such (unordered) pairs: [math]([/math][math]0[/math][math], 100), (1, 99), \ldots, (49, 51), (50, 50)[/math].If you limit x and y to be integers [math](\mathbb{Z})[/math], then you have infinitely many pairs because for all [math]x \in \mathbb{Z}[/math] there exists [math]y = (100 - x) \in \mathbb{Z}[/math]. Similarly for real numbers.If you limit [math]x[/math] and [math]y[/math] to be in the set [math]\{-40, 1, 3, 39.5, 51, 60.5, 99, 140\} [/math]then there are only 3 such (unordered) pairs; namely [math](-40, 140), (1, 99)[/math] and [math](39.5, 60.5)[/math].