C语言 输入一个任意整数,将其个位输出.

int类型所能容纳的数字位数不能超过10.我写的这个程序稍微长了点,但不受int类型容量的限制,能够处理很长的整数输入（由buffer数组的大小决定）.这程序只处理纯整数输入.有疑问尽管问. 

只学循环完全可以办到,关键是要动脑啊#includevoidmain(){\x09intn,m,max,i;\x09max=0;\x09printf("PleaseInputANumber:\n");

#includevoidmain(){inta,b,c,d;scanf("%d",&a);b=a/100;c=(a%100)/10;d=a%10;printf("%d",100*d+10*c+b);}

不对,实质上这种情况该用char类型,并判断,因为int可输入多位,例如输入11,12,14你这种算法就不能反向输出

#include#includeintmain(){inti,num,n,ans,tt;while(scanf("%d",&num)!=EOF){ans=0;n=3;//如果要求N位数反转,把n改了就

帮你写了一个,你看看：#include<stdio.h>int main(){int i = 0;int num = 

一：#includevoidmain(){inti,j;inta[5][3];for(i=0;i

可以用sprintf(str,"%d",num);将数字输出为字符串,再用strlen(str)检查其长度：intnulen(intnum){charstr[10];sprintf(str,"%d",

main(){intx;scanf("%d",&x);x=(x%10)*100+(x/10%10)*10+(x/100)printf("%d\n",x);}

#include#includevoidmain(){intnum;printf("input\t:");scanf("%d",&num);printf("%d\t%d\n",num,abs(

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

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){\x05intn,sum=0;printf("Typeaninte

#includeintmain(){inta;intb=0,c=0,d=0;scanf("%d",&a);if(a%5==0)c=1;if(a%7==0)d=1;elseif(c==1&&d==1)p

#include"stdio.h"#include"iostream"intmain(){inti,x,y=1;scanf("%d",&x);for(i=2;i

#includeintmain(void){intmax,min;inta,i;printf("Input10interger:");scanf("%d",&a);max=min=a;for(i=0;

#include<stdio.h>int main(){  int a[10],i,num=0;  float ave

voidmain(){inti,j,k=0,a,s[50];for(i=0;i

#includemain(){inti,j,N,t,k=0;intscore[100];printf("请输入整数的个数N：\n");scanf("%d",&N);printf("请输入N个整数：\n

inti,a,n;int*pl=NULL;printf("inputn:");scanf("%d",&n);if(n>0){pl=(int*)malloc(n*sizeof(int));}for(i=

#include#includevoidmain(){\x09intn,i;\x09scanf("%d",&n);\x09if(n==1)\x09{\x09\x09printf("1isnothing