试编程判断输入的正整数

main(){inta,b,num1,num2,temp;printf("请输入两个正整数:\n");scanf("%d,%d",&num1,&num2);if(num1

#includeintmain(){inta,sum=0;scanf("%d",&a);while(a){sum+=a%10;a/=10;}printf("sum=%d\n",sum);return0

DimXAsInteger这句出的毛病.你直接把它定义成整数型变量,输入3.5直接四舍五入转换为4.把它换成DimXAsSingle

#includeintmain(){intm[10],i,j=0,k;longn;scanf("%ld",&n);k=n;while(k>0){k/=10;j++;}i=j;while(i--){m[

算法一：#includemain(){intn1,n2,gcd=1,k=2;scanf("%d%d",&n1,&n2);while(k0;i--)if(n1%i==0&&n2%i==0)printf(

importjava.util.*;publicclassTest40014{publicstaticvoidmain(String[]args){Scannerin=newScanner(Syst

#include#includeintmain(){printf("请输入一个正整数：");intn1,n2;boolflag1=false,flag2=false;scanf("%d",&n1);n

#include"stdio.h"voidmain(){inti=1;intn;printf("请输入n");scanf("%d",&n);for(i=1;i再问：我知道了在FOR循环后面再加以个pr

#include"stdio.h"intmain(){\x09intn;\x09scanf("%d",&n);\x09if(n%7==0||n%5==0)\x09{\x09\x09printf("ye

#include#includeintmain(void){intn,m,i,j,t;scanf("%d%d",&n,&m);i=m>n?m:n;j=m>n?n:m;while(j){t=i%j;i=

饭要一口一口吃,作业要自己做

#include <stdio.h>#include <math.h>int main(void){    

#includeintmain(){intnum;inta;//百位intb;//十位intc;//个位scanf("%d",&num);a=num/100;//默认强制转为整型b=(int)(num

INPUT"请输入一个大于2的整数"TONFLAG=T//这个t是表示ture,flag是一个标记变量,FORI=2TON-1//这个循环i的值由2-i-1IFMOD(N,I)=0//在2-i-1这些

#include<stdio.h>main(){ int ia; scanf("%d",&ia); if(ia%5==0

算法没问题,按你的代码,调整了一下PrivateSubCommand1_Click()Dima!,b!,c!a=Val(InputBox("请输入a"))b=Val(InputBox("请输入b"))

PHP代码：$number=25;//随便写了个数.if($number%5==0&&$number%7==0){echo'yes';}else{echo'no';}就是判断对5取余和对7取余等于0.

有时间和空间要求么?简单方法如下：count=0;for(i=A;i再问：你的好像不行，这是我写的，看看怎么改一下#include#includeintmain(void){intA,B,count=

#include <iostream>void main(){    int a,b,sum,n=0; &n

main(){intd;scanf("%d",&d);if((d%3)||(d%8){printf("no!\n");}elseprintf("yes!\n");}再问：我编写了，好像此处if((d%