每日练题（py,c,cpp）.6_19,6_20

检验素数

from math import sqrt
a = int(input("请输入一个数："))
for i in range(2,int(sqrt(a))):if a%i == 0:print("该数不是素数")breakelse:    print("该数是素数")# # 1既不是素数也不是合数
# #可以用flag做标志位
# b = int(input("请输入一个数："))
# Flag = False
# for i in range(2,b):
#     if b%i == 0:
#         Flag = True# if Flag:
#     print("是合数")# else:
#     print("是素数")

#include <iostream>/* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char** argv) {int a = 0;printf("请输入一个大于1的自然数：");scanf("%d", &a);int flag = 0;for (int i = 2; i < a; i++){if (a % i == 0){flag = 1;break;}}if (flag == 0){printf("输入的数是一个素数");}else{printf("输入的数不是一个素数");}return 0;
}

逐字检查法

自解

解法1
a = "python"
b = "typhon"
list1 = list(b)
print(list1)
def judge(a,b):for i in range(len(a)):for j in range(len(a)):if a[i] == b[j]:list1[j] = Nonebreakfor i in range(len(a)):list1[i] == Noneif list1[i] != None:return Falsereturn True
if judge(a,b) == True:print("是变位词")
elif judge(a,b) == False:print("不是变位词")

解法2
a = "python"
b = "typhon"
def judge(a,b):list1 = list(a)list2 = list(b)j = len(list2)list1.sort()list2.sort()for i in range(len(list1)):if list1[i] == list2[i]:passelse:return Falsereturn True
if judge(a,b) == True:print("是变位词")
elif judge(a,b) == False:print("不是变位词")

计数比较法

 

a = "pythoe"
b = "typhon""""计数比较-自解"""
dict1 = dict()
dict2 = dict()
list1 = list(a)
list2 = list(b)for i in range(ord("a"),ord("z")+1):dict1[i] = 0for i in range(ord("a"),ord("z")+1):dict2[i] = 0def judge(a,b):for j in range(len(a)):for i in range(ord("a"),ord("z")+1):if ord(list1[j]) == i:dict1[i] = dict1[i] + 1for j in range(len(a)):for i in range(ord("a"),ord("z")+1):if ord(list2[j]) == i:dict2[i] = dict2[i] + 1for i in range(ord("a"),ord("z")+1):if dict1[i] == dict2[i]:passelse:return Falsereturn True
if judge(a,b) == True:print("是变位词")
elif judge(a,b) == False:print("不是变位词")"""计数比较-仿他解"""
def judge(a,b):c1 = [0] * 26c2 = [0] * 26for i in range(len(a)):pos = ord(a[i]) - ord('a')c1[pos] = c1[pos] + 1for i in range(len(a)):pos = ord(b[i]) - ord('a')c2[pos] = c2[pos] + 1for i in range(26):if c1[i] == c2[i]:passelse:return Falsereturn True
if judge(a,b) == True:print("是变位词")
elif judge(a,b) == False:print("不是变位词")

 

 他解

 

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鹏哥C语言

 

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