java从键盘输入一个数,是3位数就加上100,是4位数就除以100
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importjava.io.*;importjava.util.*;classMain{publicstaticintgcd(intm,intn){while(true){if((m=m%n)==0)
packagetest1;importjava.util.Scanner;publicclassTest6{publicstaticvoidmain(String[]args){Scannerinpu
importjava.util.Scanner;publicclassSum{publicstaticvoidmain(String[]args){Scanners=newScanner(System
publicstaticvoidprintNum(intn){for(inti=2;ij){if(i%j==0){m++;}}else{break;}}if(m<1){System.out.print
importjava.util.Arrays;importjava.util.Scanner;publicclassDataCompare{\x09publicstaticvoidmain(Strin
import java.util.Scanner; public class TestJT { public&nbs
a*a=b假设你先循环2到a,发现b都不能整除,这时你在循环a+1到a*a就没有意义了,因为b=a*a,所以b/(a+x)是肯定小于a的,而2到a已经循环过了不是吗?不过一般代码里面都是循环2到b/2
input "输入一个数:" to srsrc=ALLTRIM(STR(sr))lensr=LEN(src)yxsr=""FOR&
Scannerscanner=newScanner(system.in)intx,y,z,temp;System.out.println("请输入x:");x=scanner.nextInt();Sy
importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){System.out.print("请输入第一个整
//写好了#includeintmain(){boolishave=false;inta,b,num1,num2,temp;intc,GCD,LCM;//最大公约数GCD,最小公倍数LCMprintf
#include"stdafx.h"voidmain(){intnum;printf("输入一个数:");scanf("%d",&num);if(num%3==0){if(num%5==0){prin
判断从键盘输入的三个数能否构成三角形.如果是特殊的三角形(如等边、等腰、直角,等腰直角)需要能判断出来.
oolcalc(unsignedintn){if(n
clearinput"请输入一个数:"toxifx%2=1?"是奇数"else?"是偶数"endif第2题是什么?
publicclassJs{privatedoublea;privatedoubleb;privatedoublec;Js(doublea,doubleb,doublec){this.a=a;this
其实这种逻辑运算的,提现不错Java的面向对象的概念的,你直接写逻辑语句就好,首先判断等边,即if(a==b==c)System.out.print("该三角形是等边三角形");这样写下去就好了.希望
#includevoidmain(){intn;sacnf("%d",&n);if(n%27==0)printf("Yes\n");elseprintf("No\n");}
classOddSum//定义类{publicstaticvoidmain(Stringargs[])//main方法{intsum=0;//定义一个整形变量for(inti=1;i
哥们怎么又是你来?再问:您会不会的啊再答:packagetestt;importjava.util.Scanner;publicclassTest{\x09publicstaticvoidmain(S