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ptrrsn
#198
Quote:


Solusi gw:
Code:

#include <stdio.h>

char f (int r, int c, int n) {
\tif (c < 0 || c >= (1 << n) + 1 + r) return ' ';
\tif (n == 1) {
\t\tif (r + c == 0) return ' ';
\t\tif (r + c == 1) return '/';
\t\tif (c - r == 2) return '\\';
\t\treturn '_';
\t}
\tif (r >= (1 << (n - 1))) {
\t\tif (c >= (1 << n)) c -= 1 << n;
\t\tr -= 1 << (n - 1);
\t} else c -= 1 << (n - 1);
\treturn f (r, c, n - 1);
}

int main () {
\tint n;
\tbool first = true;
\twhile (scanf ("%d", &n) == 1) {
\t\tif (!first) puts ("");
\t\tfirst = false;
\t\tint r = 0;
\t\tfor (int i = (1 << n) + 1; i <= (2 << n); i++) {
\t\t\tfor (int j = 0; j < i; j++) {
\t\t\t\tprintf ("%c", f (r, j, n));
\t\t\t}
\t\t\tputs ("");
\t\t\tr++;
\t\t}
\t}

return 0;
}
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