3. data comparison size

Here we mainly discuss the implementation of algorithms, which seem simple. In fact, different algorithms have different implementation ideas. Assume that a common example is to input three integers, a, B, and c, in ascending order. Method 1: Comparison between a, B, and c, there are 6 combinations, using if... else if implementation, as follows: # include <stdio. h> int main () {int a, B, c; scanf ("% d", & a, & B, & c ); if (a <= B & B <= c) printf ("% d \ n", a, B, c ); else if (a <= c & c <= B) printf ("% d \ n", a, c, B ); else if (B <= a & a <= c) printf ("% d \ n", B, a, c ); else if (B <= c & c <= a) printf ("% d \ n", B, c, ); else if (c <= a & a <= B) printf ("% d \ n", c, a, B); else if (c <= B & B <= a) printf ("% d \ n", c, B, a); return 0 ;}method 2: I think there are too many branches above. a and B are compared first. If a> B, they are exchanged. Then a and c are compared. If a> c, they are exchanged. Finally, B and c are compared, if B> c, switch. # Include <stdio. h> int main () {int a, B, c, t; scanf ("% d", & a, & B, & c ); if (a> B) {t = a; a = B; B = t;} if (a> c) {t = a; a = c; c = t ;} if (B> c) {t = B; B = c; c = t;} printf ("% d \ n", a, B, c ); return 0;} method 3: set the current minimum value x and the maximum value z. The initial value is a. With the comparison, the minimum value and the maximum value are updated slowly. To the final maximum and minimum values. If you know the minimum and maximum values, you can calculate the Middle Value y. # Include <stdio. h> int main () {int a, B, c, x, y, z; scanf ("% d", & a, & B, & c); // min x; max zx = a; if (B <x) x = B; if (c <x) x = c; z =; if (B> z) z = B; if (c> z) z = c; y = a + B + c-x-z; printf ("% d \ n", x, y, z); return 0;} additional question: enter a positive integer n <= 30, output an inverted triangle of n layers. For example, when n = 5, the output is as follows: *************************. My implementation is as follows: # include <stdio. h> int main () {int I, t, j, k; scanf ("% d", & I); for (t = I; t> = 1; t --) {// output two space characters first for (k = I-t; k> 0; k --) {printf ("");} // output the "*" character for (j = 2 * t-1; j> = 1; j --) {printf ("*");} // line feed printf ("\ n");} return 0 ;}