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柚茶_Rola
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题解 | #称砝码#
n = int(input()) m = list(map(int,input().split())) x = list(map(int,input().split())) amount = [] for i in range(n): for j in range(x[i]): ...
2024-10-21
0
41
题解 | #字符串加密#
key = input() s = input() A = 'abcdefghijklmnopqrstuvwxyz' kkey = '' for i in key: if i not in kkey: kkey += i for i in A: if i not ...
2024-10-20
0
43
题解 | #密码截取#
while True: try: s = input() p = 0 for i in range(len(s)): for j in range(len(s)-1,i-1,-1): i...
2024-10-19
0
32
题解 | #字符串加解密#
# 原字符 A = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz' # 变换后的字符 B = '1234567890bcdefghijklmnopqrstuvwxyzaBCDEFGHIJKLMNOPQRSTUVWX...
2024-10-19
0
53
题解 | #查找兄弟单词#
while True: try: s = input().split() n = int(s[0]) ls = s[1:-2] x = s[-2] k = int(s[-1]) lst = []...
2024-10-19
0
43
题解 | #字符串排序#
# 2024年10月19日 下午18:35 # A Famous Saying: Much Ado About Nothing (2012/8). # A aaAAbc dFgghh: iimM nNn oooos Sttuuuy (2012/8). s = input() a = '' ...
2024-10-19
0
39
题解 | #合唱队#
# 计算每个人左边比他矮的人数,最小值是1 def f(ls): dp = [1]*len(ls) for i in range(len(ls)): for j in range(i+1): if ls[j] < ls[i]: ...
2024-10-19
0
43
题解 | #密码验证合格程序#
def f1(x1): if len(x1) > 8: return True return False def f2(x2): up = 0 lw = 0 num = 0 ot = 0 for i in x2: ...
2024-10-19
0
35
题解 | #二维数组操作#
while True: try: m,n = list(map(int,input().split())) x1,y1,x2,y2 = list(map(int,input().split())) x = int(input()) ...
2024-10-18
0
46
题解 | #密码强度等级#
s = input() sc = 0 #1 密码长度 if 0 <= len(s) <= 4: sc += 5 elif 5 <= len(s) <= 7: sc += 10 elif len(s) >= 8: sc += 25 up = 0 ...
2024-10-18
0
47
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