思路是a是逐渐增大的一个数，所以只要a发生变化，a的计算结果就一定没出现过，所以count+1.n = int(input())tem = -1count = 0for i in range(1,n+1):a = i//2 + i//3 + i//5if tem != a:count += 1tem = aprint(count)...

m,x,y = map(int, input().split(" "))def judge(m,a):if a>m:res = "Cong"elif a==m:res = "Ping"else:res = "Gai"return reslist1 = []for a in range(10...

首先要判断输入的数是否为回文如（1,11,121,123321,1234321）,如果是则直接输出，不是再进入循环。a = input()count = 0flag = 0if a == a[::-1]:print("%s is a palindromic number." % a)flag =1if flag==0:while 1:...

19分，有一个测试点为非零返回import mathl,k = map(int,input().split(" "))n = input()for i in range(l-k+1):flag = 1num = int(n[i:i+k])if num%2 == 0 and num != 2:flag = 0elif num...

# 两个分数的值可以大于1，所以要使用while循环，每次循环不断递增，直到大于最大的那个分数import matha,b,k = map(str,input().split(" "))k = int(k)n1,m1 = map(int,a.split("/"))n2,m2 = map(int,b.split("/"))res1 = n1

第4、5测试点超时，大佬们有空看一下~n = int(input())having_list = []for i in range(n):#将每一对情侣放入元组中，再将元祖放入列表list1 = input().split(" ")having_list.append((list1[0],list1[1]))m = int(input())ipt_...

str1 = input()ipt = list(input())new_list = []if str1 == "C":a = ipt[0]ipt.append("0")count = 1for i in ipt[1:]:if i == a:count += 1else:...

最后一个测试点答案错误，得19分，老哥们看看是为啥P,M,N = map(int, input().split(" "))dict1 = {}for i in range(P):ipt_list = input().split(" ")ipt_list[1] = int(ipt_list[1])if ipt_list[0] not in dict1.key.