数据结构/C++：AVL树

执行时长

def min_execution_time(n, size, tasks):
    a = 0
    ans = size

    i = 0
    while i < size:
        tmp = tasks[i]
        a += tmp
        if a < n:
            a = 0
        else:
            a -= n

        i += 1

    ans += a // n

    if a % n != 0:
        ans += 1

    return ans

# 读取输入
n = int(input())
size = int(input())
tasks = list(map(int, input().split()))

# 输出结果
result = min_execution_time(n, size, tasks)
print(result)

万能字符单词拼写

def count_valid_words(words_count, word_arr, chars):
    char_arr = [0] * 26
    any_num = 0
    res = 0

    for c in chars:
        if c == '?':
            any_num += 1
        else:
            char_arr[ord(c) - ord('a')] += 1

    for word in word_arr:
        word_split = [0] * 26

        for c in word:
            word_split[ord(c) - ord('a')] += 1

        need_any_num = sum(max(0, word_split[i] - char_arr[i]) for i in range(26))

        if need_any_num <= any_num:
            res += 1

    return res

if __name__ == "__main__":
    words_count = int(input())
    word_arr = [input() for _ in range(words_count)]
    chars = input()

    result = count_valid_words(words_count, word_arr, chars)
    print(result)

来自异国的客人

有位客人来自异国，在该国使用m进制计数。该客人有个幸运数字n(n<m)，每次购物时，其总是喜欢计算本次支付的花费(折算为异国的价格后)中存在多少幸运数字。问：当其购买一个在我国价值k的产品时，其中包含多少幸运数字？

输入描述
第一行输入为 k, n, m。

其中：

k 表示 该客人购买的物品价值（以十进制计算的价格）

n 表示 该客人的幸运数字

m 表示 该客人所在国度的采用的进制

输出描述
输出幸运数字的个数，行末无空格。当输入非法内容时，输出0

# coding:utf-8
#JSRUN引擎2.0，支持多达30种语言在线运行，全仿真在线交互输入输出。 
import functools
import sys
from collections import Counter, defaultdict
import copy
from itertools import permutations
import re
import math
import sys
from queue import Queue
 
k=0
n=0
m=0
flag = False
try:
    params = [int(x) for x in input().split(" ")]
    k= params[0]
    n = params[1]
    m = params[2]
    if(k <= 0 or n <= 0 or m <= 1 or n >= m):
        print(0)
        flag = True
except :
    print(0)
    flag = True
 
if(not flag) :
    result = 0
    while(True):
        if(k<=0) :
            print(result)
            break
        else :
            if (k % m == n) :
                result+=1
                k //= m
            else:
                k //= m
             

求最多可以派出多少支团队

n = int(input())
v = list(map(int, input().split()))
t = int(input())
v.sort(reverse=True)
left = 0
right = n - 1
res = 0
while left < right:
    if v[left] >= t:
        res += 1
        left += 1
    else:
        while left < right:
            if v[left] + v[right] >= t:
                left += 1
                right -= 1
                res += 1
                break
            right -= 1
print(res)                              

最长子字符串的长度（一）

import sys
 
for line in sys.stdin:
    input_str = line.strip()
    char_o_nums = input_str.count('o')
    if char_o_nums % 2 == 0:
        print(len(input_str))
    else:
        print(len(input_str) - 1)

机场航班调度程序

 
class Main:
    @staticmethod
    def schedule(mv):
        ost = ""
        for key, value in sorted(mv.items()):
            for v in sorted(value):
                ost += key + format(v, '04d') + ","
        ans = ost[:-1]  # Remove the trailing comma
        return ans
 
    @staticmethod
    def main():
        s = input()
        s = s.replace(",", " ")
        tokens = s.split()
        mv = {}
        for token in tokens:
            key = token[:2]
            value = int(token[2:])
            if key not in mv:
                mv[key] = []
            mv[key].append(value)
        print(Main.schedule(mv))
 
Main.main()

数的分解

 
def find_consecutive_sum(n):
    result = []
    start = 1
    end = 2
    while start < end:
        current_sum = (start +end) * (end - start +1)//2
        if current_sum == n:
            result.append(list(range(start,end + 1)))
            start += 1
        elif current_sum < n:
            end += 1
        else:
            start += 1
    if result:
        min_length = min(len(seq) for seq in result)
        for seq in result:
            if len(seq) == min_length:
                return f"{n}={'+'.join(str(i)for i in seq)}"
    else:
        return "N"
n = int(input())
print (find_consecutive_sum(n))