输入一个小于五的整数,求它是几位数

#includevoidmain(){inti,n,z,f,x;printf("请输入一共要处理几个数：");scanf("%d",&n);for(i=z=f=0;i0)z++;elseif(x

#includelongf(longintx){inta[100];inti;for(i=1;;i++){a[i]=x%10;x=x/10;if(x==0)break;}return(i);}void

varn:longint;beginreadln(n);ifnmod7=0thenwriteln('yes'){除以7取余,是否等于0,等于则输出yes,否则输出no}elsewriteln('no'

给你写了个,运行通过,你看看吧,记得采纳哦O(∩_∩)O~#includeintmain(){\x09intnum,i=0;\x09printf("pleaseinputanumble:");\x09

#includeintmain(){\x05inti,k=0;\x05for(scanf("%d",&i);i;i=i/10)\x05\x05k++;\x05printf("是%d位数",k);\x0

这样才完整：判断不出错.N-1省去了许多判断,提高了效率.PrivateSubCommand1_Click()DimNAsIntegerDimiAsIntegerIfIsNumeric(Text1.T

你写的实在是太繁琐了.这个是不可取的.你需要的这两个功能其实是可以一起实现的,虽然一楼说的很对,当我觉得应该不是你需要的那一种方案.#includemain(){inti=0,k;inta[10;pr

printf("输入一个小于1000的整数x",x)；x不需要,改成printf("输入一个小于1000的整数:")；printf("输入的数据不符合要求,重新输入一个小于1000的整数x",x)同理

#include#includeintmain(void){intn,sum=0,count=0,s;printf("输入一个整数:");scanf("%d",&n);while(n){s

请输入:100357111317192329313741434753596167717379838997Pressanykeytocontinue#include#includeintmain(){\

请输入:100357111317192329313741434753596167717379838997Pressanykeytocontinue#include#includeintmain(){

#include<stdio.h>#define N 5int main(){    int len(int

#includeintmain(){inti=1;longnum;//int的范围是-32768~32767,这里要用long型才够longn;//复制numinta,b,c,d,e;scanf("%

scanf("%d",&a);再问：这是怎么回事？？求解T^T再答：#includevoidmain(){inta,b,c,d,e,f,g;scanf("%d",&a);b=a/1000;c=a-(b

/*从键盘上输入一个任意位数的正整数,判断它是几位数,并逆序输出该数*/#include"stdio.h"intmain(){intnum,temp,i=0;printf("请输入一个数字\n");s

具体代码如下:#includeintmain(){intn,i=0;printf("Entern:");scanf("%d",&n);while(n){printf("%d",n%10);n/=10;

#include#includevoidmain(void){inti,j,k,f,z;scanf("%d",&i);if(i

#includeusingnamespacestd;voidmain(){intn;cin>>n;if(n%2==0)cout

#include<stdio.h>#include<math.h>int main(){   int x,y; &n

#include#defineMAXN20intmain(){intn,t,k=0;intia[MAXN];printf("请输入一个整数:");scanf("%d",&n);while(t=n%10