add basic 1121-1125

PAT (Basic Level) Practice （中文）/1121_祖传好运 (15).py

@@ -0,0 +1,12 @@
+def is_lucky(num_str):
+    result, val = True, 0
+    for i in range(0, len(num_str)):
+        val = val * 10 + int(num_str[i])
+        result &= val % (i + 1) == 0
+    return result
+
+
+input()
+num_str_arr = input().split()
+for num_str in num_str_arr:
+    print("Yes" if is_lucky(num_str) else "No")

PAT (Basic Level) Practice （中文）/1122_找奇葩 (20).py

@@ -0,0 +1,9 @@
+input()
+arr = map(int, input().split())
+# filter out even numbers
+arr = list(filter(lambda v: v % 2 == 1, arr))
+result = arr[0]
+# xor remove odd number pair
+for i in range(1, len(arr)):
+    result ^= arr[i]
+print(result)

PAT (Basic Level) Practice （中文）/1123_舍入 (20).py

@@ -0,0 +1,26 @@
+from decimal import *
+
+
+def round_number(method, number, precision):
+    context = getcontext()
+    context.prec = 300  # 精度位数一定要大
+    if method == "1":
+        context.rounding = ROUND_HALF_UP
+    elif method == "2":
+        context.rounding = ROUND_DOWN
+    elif method == "3":
+        context.rounding = ROUND_HALF_EVEN
+
+    d = Decimal(number).quantize(Decimal("0." + "0" * precision))
+    if d == 0:
+        d = abs(d)
+    ans = str(d)
+    if ans[0] == "+":
+        ans = ans[1:]
+    return ans
+
+
+n, d = map(int, input().split())
+for i in range(n):
+    op, num = input().split()
+    print(round_number(op, num, d))

PAT (Basic Level) Practice （中文）/1124_最近的斐波那契数 (20).py

@@ -0,0 +1,18 @@
+fib = [0, 1]
+
+count = 0
+while fib[-1] <= 10**8:
+    new = fib[-1] + fib[-2]
+    fib.append(new)
+
+v = int(input())
+ans, diff = -1, 10**8
+for i in range(len(fib)):
+    if fib[i] >= v:
+        if abs(fib[i] - v) < diff:
+            ans = fib[i]
+        break
+    else:
+        ans = fib[i]
+        diff = abs(fib[i] - v)
+print(ans)

PAT (Basic Level) Practice （中文）/1125_子串与子列 (25).cpp

@@ -0,0 +1,30 @@
+#include <iostream>
+#include <vector>
+using namespace std;
+
+int main() {
+	ios::sync_with_stdio(false);
+	string s, p; 
+	cin >> s >> p;
+
+	pair<int, string> ans{10e5+10, ""};
+	// brute force match p in s and find the shortest substring
+	for (int i = 0; i < s.size(); i++) {
+		if (s[i] != p[0]) continue;
+		int l = i, r = l+1;
+		int k = 1;
+		for (; r < s.size() && k != p.size(); r++) {
+			if (s[r] == p[k]) k++;
+			if (k == p.size()) break;
+		}
+
+		int len = r-l+1;
+		if (k == p.size() && len < ans.first) {
+			ans.first = len;
+			ans.second = s.substr(l, len);
+		}
+	}
+
+	cout << ans.second << endl;
+	return 0;
+}