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Commitec8ecd4dc19854007428761c330303854478201f (patch)
Author
 Syndamia <kamen@syndamia.com>
2025-11-06 19:13:54 +0200
Committer
 Syndamia <kamen@syndamia.com>
2025-11-06 19:15:07 +0200
Files662f55a328b60a70273a5e6fc564b1511da906cf /Week06/Exercise05.cpp
upp-2025-solutions-ec8ecd4dc19854007428761c330303854478201f.tar
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Parent9242ebc4c69f2c1f550ddcc0e9235f3f409554c7 (diff)
feat(Week06): Add exercises and solutions
1 files changed, 31 insertions, 0 deletions
-rw-r--r--Week06/Exercise05.cpp31
diff --git a/Week06/Exercise05.cpp b/Week06/Exercise05.cpp
new file mode 100644
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--- /dev/null
+++ b/Week06/Exercise05.cpp
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+#include <iostream>
+
+int main() {
+ double x, y;
+ std::cin >> x >> y;
+
+ bool below_top_left_corner = (x > -4) && (y < 2);
+ bool above_bottom_left_corner = (x > -4) && (y > -1);
+
+ /* Уравнение на правата:
+ * y = mx + b
+ * Където m е наклон.
+ * m = (промяна в y) / (промяна в x)
+ * ОБАЧЕ знакът на m е отрицателен, когато наклонът е на дясно (от горе в ляво до долу в дясно)
+ * Правата "отива нагоре" с 1 и на дясно с 1, следвателно m = 1/1 = 1
+ * Когато x = 0, тогава y = 1, следователно b = 1
+ *
+ * Уравнението за нашата права е:
+ * y = x + 1
+ *
+ * Точка е над правата, когато y > x + 1 (и под правата когато y < x + 1)
+ */
+ bool is_above_line = y > x + 1;
+
+ if (below_top_left_corner && above_bottom_left_corner && is_above_line)
+ std::cout << "Inside" << std::endl;
+ else
+ std::cout << "Outside" << std::endl;
+
+ return 0;
+}