add basic 1116-1120

Author: tiny656

PAT (Basic Level) Practice （中文）/1116_多二了一点 (15).py

@@ -0,0 +1,9 @@
+s = input()
+if len(s) % 2 == 1:
+	print('Error: %d digit(s)' % len(s))
+else:
+	a, b = int(s[:len(s)//2]), int(s[len(s)//2:])
+	if b - a == 2:
+		print('Yes: %d - %d = 2' % (b, a))
+	else:
+		print('No: %d - %d != 2' % (b, a))

PAT (Basic Level) Practice （中文）/1117_数字之王 (20).py

@@ -0,0 +1,38 @@
+def func(arr):
+	is_all_single = True
+	mem = []
+	for i, v in enumerate(arr):
+		if v >= 10: is_all_single = False
+
+		r = 0 if v == 0 else 1
+		while v != 0:
+			r *= (v % 10) ** 3
+			v //= 10
+
+		v, r = r, 0
+		while v != 0:
+			r += v % 10
+			v //= 10
+
+		mem.append((i, r))
+	if is_all_single: return
+	else:
+		for i, v in mem:
+			arr[i] = v
+		func(arr)
+
+n1, n2 = map(int, input().split())
+arr = [i for i in range(n1, n2+1)]
+
+func(arr)
+max_count = 0
+ans = []
+for v in set(arr):
+	cnt = arr.count(v)
+	max_count = max(max_count, cnt)
+	ans.append((v, cnt))
+
+ans = filter(lambda x: x[1] == max_count, ans)
+ans = sorted(ans, key=lambda x: x[0])
+print(max_count)
+print(' '.join(map(lambda x: str(x[0]), ans)))

PAT (Basic Level) Practice （中文）/1118_如需挪车请致电 (20).py

@@ -0,0 +1,21 @@
+pinyin = ['ling', 'yi', 'er', 'san', 'si', 'wu', 'liu', 'qi', 'ba', 'jiu']
+
+def get_number(s):
+	if 'sqrt' in s:
+		n = int(s[4:])
+		return int(n ** 0.5)
+	
+	if s in pinyin:
+		return pinyin.index(s)
+	
+	if '^' in s:
+		n, m = map(int, s.split('^'))
+		return int(n ** m)
+	
+	return int(eval(s))
+
+phone = []
+for i in range(11):
+	s = input()
+	phone.append(get_number(s))
+print(''.join(map(str, phone)))

PAT (Basic Level) Practice （中文）/1119_胖达与盆盆奶 (20).py

@@ -0,0 +1,28 @@
+# construct peak array from left and right, then find the max of each element in the two arrays
+# stands for the max height, then add 1 and multiply 100 is required the volume of milk
+n = int(input())
+arr = list(map(int, input().split()))
+
+left, right = [1], [1]
+for i in range(1, n):
+	if arr[i] > arr[i-1]:
+		left.append(left[i-1] + 1)
+	elif arr[i] == arr[i-1]:
+		left.append(left[i-1])
+	else:
+		left.append(1)
+
+for i in range(n-2, -1, -1):
+	if arr[i] > arr[i+1]:
+		right.append(right[n-i-2] + 1)
+	elif arr[i] == arr[i+1]:
+		right.append(right[n-i-2])
+	else:
+		right.append(1)
+
+right = list(reversed(right))
+
+ans = 0
+for i in range(n):
+	ans += (max(left[i], right[i]) + 1) * 100
+print(ans)

PAT (Basic Level) Practice （中文）/1120_买地攻略 (25).cpp

@@ -0,0 +1,41 @@
+#include <iostream>
+#include <vector>
+#include <algorithm>
+using namespace std;
+
+typedef long long int64;
+
+// Solution 1: O(n^2), 遍历所有可能的区间并统计合法的区间数
+int Solution1(const vector<int64> &sum, int64 n, int64 m) {
+	int64 cnt = 0;
+	for (int i = 0; i <= n; i++) {
+		for (int j = i+1; j <= n; j++) {
+			if (sum[j]-sum[i] <= m) cnt++;
+		}
+	}
+	return cnt;
+}
+
+// Solution 2: O(nlogn), 二分查找, 从后往前遍历, 对于每个i, 二分查找第一个pos, 使得sum[pos] >= sum[i]-m
+int Solution2(const vector<int64> &sum, int64 n, int64 m) {
+	int64 cnt = 0;
+	for (int i = n; i >= 1; i--) {
+		int pos = lower_bound(sum.begin(), sum.end(), sum[i] - m) - sum.begin();
+		cnt += i - pos;
+	}
+	return cnt;
+}
+
+int main() {
+	ios::sync_with_stdio(false);
+	int64 n, m, v;
+	cin >> n >> m;
+	vector<int64> sum(n+1, 0);
+	for (int i = 1; i <= n; i++) {
+		cin >> v;
+		sum[i] = sum[i-1] + v;
+	}
+	// cout << Solution1(sum, n, m) << endl;
+	cout << Solution2(sum, n, m) << endl;
+	return 0;
+}