这里用到了三个函数: #判断是否为数字:str.isdigit()#是否为字母:str.isalpha()#是否为空格:str.isspace() def tongji(str): alpha = 0 number = 0 space =0 qt = 0 for i in range(len(str)): #或者for i in str: if str[i].isalpha(): #接上一句改为:i.isalpha() alpha += 1 elif str[i].isdigit(): numb
以下代码:input验证码框,输入非数字或非12位时,红框提示;每4位加一个空格 //input验证码框,输入非数字或非12位时,红框提示;每3位加一个空格 $(".text").focus(function(){ $(this).keyup(function(){ var num = /^[0-9]*$/; var verifyValue = $(this).val(); var deleteLogo = $(".delete-logo");//删除的小叉号 ve
输入任意一个大写字母,生成金字塔图形 def GoldTa(input): L = [chr(i) for i in range(65, 91)] # 大写字母A--Z idA = 65 # 从A开始 # ord()函数将字母转换为Unicode数值 idInput = ord(input) num = idInput - idA + 1 # 输入的字符个数 tempResult = "" for C in range(0, num): for C1 in range(0, C): #
import java.util.Scanner; /** * 功能:用位运算,判断一个数是否为2的n次方. * 思路:用1做移位操作,然后判断移位后的值是否与给定的数相同. */ public class Main3 { public boolean isPower(int n) { if (n < 1) { return false; } int m = 1; while (m < n) { m = m << 1; } if (m == n) { return true; }
package a; public class ShuZi { int m; public int getM() { return m; } public void setM(int m) { this.m = m; } public void shu() { System.out.println("输入的数字是:"+m); if(m>99999) { System.out.println("The number is too large"); } else
注1:10.102. 可以替换成其他的,或者手动输入. 注2:minion-可以替换为其他的,或者手动输入. 代码如下(python3): import re # list = [] list1 = [] list2 = [] list3 = [] list4 = [] with open('srcIp.txt') as fip,open('desIP','w') as f: for data in fip.readlines(): # data = re.sub(r'[\s|,]', '\n'
Sub 图片旋转任意角度() Dim sha As Shape, isa As InlineShape Static s As Integer Application.ScreenUpdating = 0 s = InputBox("请输入要旋转的角度:", "图像旋转", s) If ActiveDocument.Range.InlineShapes.Count > 0 Then For n = 1 To ActiveDocument.Range.Inlin
