# 解法一 n = int(input()) ans = 0 for i inrange(1, n): x = str(i) for j inrange(len(x)-1): if x[j] > x[j+1]: break else: ans += 1 print(ans)
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# 解法二 import sys
num = 0 n = int(input()) for i inrange(1, n+1): c = i b = sys.maxsize # 获取最大整数值sys.maxsize while c > 0: a = c % 10 if a > b: break b = a c //= 10# 使用整数除法 if c == 0: num += 1 print(num)
n = int(input()) s = list(map(int,input().split())) max_num = 0 for i inrange(n-1): for j inrange(i+1,n): max_num = max(max_num,abs(j-i) + abs(s[j]-s[i])) print(max_num)
# 解法一:使用字典来计算各个字符出现的次数,再单独使用一个循环来判断是否有不唯一的字符出现 c = input().strip() n = len(c) count = {} # 初始化计数器字典 for i inrange(n): if c[i] in count: count[c[i]] += 1 else: count[c[i]] = 1
# 检查是否有重复的字符 ifany(value > 1for value in count.values()): print("NO") else: print("YES")
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# 解法二:set(s) 将字符串 s 转换成一个集合,集合中的元素是唯一的,这意味着所有重复的字符都会被移除 s=input() iflen(set(s))==len(s): print('YES') else: print('NO')
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# 解法三:类似解法一,换了一种表达方式,使用列表来存储数据 s = input() s = s.upper() d = [] for i in s: if i notin d: d.append(i) else: print("NO") break else: print("YES")
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# 解法四:从第一个字符开始判断字符串中有几个该字符出现 # 使用到了str.count(sub)方法来计算子字符串sub在字符串s中出现的次数 s = input() for word in s: if s.count(word) != 1: print("NO") break if word == s[-1]: print("YES")
a=input() b=input() c=0 for i in a: if a.count(i)==b.count(i): continue else: print(i) c=1 break for i in b: if c==1: break else: if b.count(i)==a.count(i): continue else: print(i)
defselection_sort(arr): n = len(arr) for i inrange(n): min_idx = i for j inrange(i + 1, n): if arr[j] < arr[min_idx]: min_idx = j arr[i], arr[min_idx] = arr[min_idx], arr[i] return arr
n = int(input()) arr = list(map(int, input().split()))
defquick_sort(arr): iflen(arr) <= 1: return arr pivot = arr[len(arr) // 2] left = [x for x in arr if x < pivot] middle = [x for x in arr if x == pivot] right = [x for x in arr if x > pivot] return quick_sort(left) + middle + quick_sort(right)
# 比较左右两部分的元素,将较小的元素添加到结果中 while i < len(left) and j < len(right): if left[i] < right[j]: result.append(left[i]) i += 1 else: result.append(right[j]) j += 1
# 将剩余的元素添加到结果中 result.extend(left[i:]) result.extend(right[j:]) return result