将键盘上输入的任意正整数逆转 如1234,输出4321
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看vari,j,n,t:longint;a:array[1..1000]oflongint;beginwhilenoteolndobegininc(n);read(a[n]);end;i:=1;j:=
intmain(){intn,n1,n2,n3,n4;cin>>n;n1=n/1000;n2=(n%1000)/100;n3=(n%100)/10;n4=n%10;n=n1+10*n2+100*n3+
int交换函数(intX){int新十位数=X用10取余;int新个位数=(X-[X用10取余])/10;交换函数=10×新十位数+新个位数;输出“交换函数”}说明:输入的数据为X.
#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}
#includemain(){inta;printf("ÊäÈëÒ»¸öÕý&
#includeintmain(){intx,y,a;scanf("%d",&x);scanf("%d",&y);a=x*y;printf("Theaverageis%d\n",a);return0;
shft+K،shft+P؛shft+L/标符是shft+qwerasx(分别每一个字母对应一种符号,你自己试试看看吧)
#include#includevoidmain(){inta,b,c;//定义三角形三边长scanf("%d%d%d",&a,&b,&c);//输入三边长,为整形doublep=(a+b+c)/2.
方法一://用数学函数#include#includevoidmain(){inta;scanf("%d",&a);printf("%d\n",abs(a));}方法二://判断#includevoi
#includeintmain(){intnumPos=0,numNeg=0,numZeros=0,num,in;printf("Pleasekeyinthenumberofintegers:\n")
最大公约数:intGcd(inta,intb){if(a%b==0)returna;return(b,a%b);}最小公倍数:intGbs(inta,intb){returna*b/Gcd(a,b);
import java.util.*;public class Demo{ public static
Inputofany4Numbersonthekeyboard
voidmain(){inti,j,k=0,a,s[50];for(i=0;i
#includevoidmain(){floata=0,b=0,c=0;chard;printf("输入两个数:\n");scanf("%f%f",&a,&b);getchar();printf("选
CLEARaccept"请输入N值:"TONs=1FORi=1TOVAL(n)IFMOD(i,2)=0s=s*iENDIFENDFOR?"1——N中的所有偶数的积为:"+ALLTRIM(STR(s))
#includeintmain(void){intn=0;intnum=0;charch;printf("Endtoinput\'a\'.\n");printf("Pleaseinputnumber:
#include "stdio.h"int main() {int n, sum=0, i=0, max,&
#include"stdio.h"voidmain(){intNum=0;printf("请输入数字:");scanf("%d",&Num);intArray[100];intn_Num=0;//记录
提供一个思路,不用照抄,没有验证过的如一个数2268,就是4+4+36+64=108用个循环取每个数出来,平方再相加X就是这个数,Y是累加数,I是循环次数fori=1tolen(x)y=y+mid(x