1 input()函数来实现用户输入,程序在等待输入的时候会终止,获取用户的输入后继续 message = input("tell me something,and I will repreat it back to you:") print(message) 2 使用int来获取数值输入 age = input("How old are you?") age = int(age) print(age) 3 求模运算符 number = input("En
num=input("please input the num: ") print "the number your input is: "+str(num) def checktype(num): if isinstance(num,(int,long,float,complex)):##isinstance 是判断是否是一个类的实例,如果是返回true print str(num)+" type is "+type(num).__name__
package seven_topic; import java.util.*; public class p_19_1 { public static void main(String[] args) { @SuppressWarnings("resource") Scanner reader = new Scanner(System.in); System.out.println("请输入学生学号: "); String a = reader.next(); S
末尾没有目的地的出租车,污点证人禁止入内!!! 不同的尝试有不同的方法 关于int类型的判断,我尝试了这么一个方法,可行,只是笨 正则表达式我没有搞清楚,没办法给大家讲解,欢迎各位明白人讲解,或者是我搞清楚弄明白了,再在此贴给大家解说 判断int类型的本方法,↓↓↓↓ Console.WriteLine("请输入:"); var x = Console.ReadLine(); try { long i = Convert.ToInt64(x); Console.WriteLine(&q
import java.util.*; class Example2_5 { public static void main(String args[]) { int start=0,end,middle; int a[] = {12,45,67,89,123,-45,67}; int N = a.length; for(int i=0; i<N; i++) { //选择法排序数组 for(int j = i+1; j < N;j++){ if(a[j] < a[i]){ int t =
private Timer timer = new Timer(); private final long DELAY = 1000; // in ms EditText editTextStop = (EditText) findViewById(R.id.editTextStopId); editTextStop.addTextChangedListener(new TextWatcher() { @Override public void beforeTextChanged(CharSeq