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")
他解
鹏哥C语言
、