从键盘输入一个五位数,如果该数是奇异魔幻数,输入"Yes",否则输出"No".

设原来的最高位为a,后四位为b,则原数=10000a+b,新数=10b+a,由已知10000a+b=10b+a+11106,9999a=9b+11106,1111a=b+1234,b=1111(a-1

packagetest1;importjava.util.Scanner;publicclassTest6{publicstaticvoidmain(String[]args){Scannerinpu

#include#includeintmain(){\x09doublex;\x09scanf("%lf",&x);\x09printf("%lf\n",sqrt(x));\x09return0;}

可参考如下scanf("%d",&num);printf("%d\n",num/100);printf("%d\n",num%10);---------------------------------

#includeintmain(){intkey,index;intarr[10]={1,2,3,4,5,6,7,8,9,10};printf("输入整数Key:\n");scanf("%d",&ke

前面三位数是个完全平方数,那么设它为x^2,后面两个都相同,设为10a+a=11a.这个数表示为A：A=100x^2+11a此数可被99整除,必然能被11整除,11a已经可被11整除了,所以100x^

#includeintmain(){inta;intsum=0;scanf("%d",&a);if(a

我有个matlab的代码,闲来无事就你这个题编的,供你参考吧.a=input('pleaseinputanumber:');j=1;fori=1:9ifa-j

#includeintmain(){intnum;inta,b,c;scanf("%d",&num);a=num/100;b=(num%100)/10;c=num%10;printf("a=%d,b=

fpc:varn:string;a,b,i:longint;beginread(n);b:=0;a:=length(n);fori:=1toadob:=ord(n[i])-ord('0')+b;wri

OptionBase1PrivateSubCommand1_Click()n=InputBox("请输入一个正整数")Fori=1ToLen(n)Sum=Sum+Val(Mid(n,i,1))Next

#includeusingnamespacestd;intmain(){char*str=newchar[255];cin>>str;cout再问：看不等，我刚学编程再答：把输入的数字放到一个指针数组

inputaifmod(a/2)=0?a/2+3endif

这要有数学的基本常识..奇数被2除余1偶数被2整除..所以假设这个数位xif(x%2==1)//xmod2余1输出奇数；else输出偶数；

#include"stdafx.h"voidmain(){intnum;printf("输入一个数:");scanf("%d",&num);if(num%3==0){if(num%5==0){prin

/*#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0&&x%5==0)printf("该数字能同时被3,

oolcalc(unsignedintn){if(n

#includeintmain(){inta[10],i,b;for(i=0;i

clearinput"请输入一个数："toxifx%2=1?"是奇数"else?"是偶数"endif第2题是什么?

clearinput"请输入一个数："toxa=0b=9x=alltrim(str(x))k=len(x)fori=1tokm=substr(x,i,1)n=int(val(m))ifn>aa=nen