【算法面试题】-04

执行时长

def min_execution_time(n, size, tasks):a = 0ans = sizei = 0while i < size:tmp = tasks[i]a += tmpif a < n:a = 0else:a -= ni += 1ans += a // nif a % n != 0:ans += 1return 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] * 26any_num = 0res = 0for c in chars:if c == '?':any_num += 1else:char_arr[ord(c) - ord('a')] += 1for word in word_arr:word_split = [0] * 26for c in word:word_split[ord(c) - ord('a')] += 1need_any_num = sum(max(0, word_split[i] - char_arr[i]) for i in range(26))if need_any_num <= any_num:res += 1return resif __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 Queuek=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 = Trueif(not flag) :result = 0while(True):if(k<=0) :print(result)breakelse :if (k % m == n) :result+=1k //= melse: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 += 1left += 1else:while left < right:if v[left] + v[right] >= t:left += 1right -= 1res += 1breakright -= 1
print(res)                              

最长子字符串的长度（一）

import sysfor 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:@staticmethoddef 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 commareturn ans@staticmethoddef 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 = 1end = 2while start < end:current_sum = (start +end) * (end - start +1)//2if current_sum == n:result.append(list(range(start,end + 1)))start += 1elif current_sum < n:end += 1else:start += 1if 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))