c语言输出一个整数交换位数

intmain(){intinput;inti;intn1,n2,n3;scanf("%d",&input);for(i=100;i

inta,i;a=245;intb[3];//存个十百位for(i=0;i=0;i--)printf("%d",b[i]);//输出百十个位

#includemain(){inti=0,s=0,n;scanf("%d",&n);while(n>0){s+=n%10;i++;n/=10;}printf("%d%d\n",s,i);}

#includemain(){inta,b,x;scanf("%d%d",&a,&b);x=a;a=b;b=x;printf("a=%d,b=%d\n",a,b);}

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

#includemain(){inti,n=0,num[100];chars;printf("请输入数字\n");for(i=1;i

用%（求余）.比如你输入的是321321%100=21用321-(321%100)再除以100即可得：（321-(321%100)）100=3同样,用21%10=1再除以10即可得：（（321%100

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. 

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

假设4位分别为a,b,c,d运用穷举法判断abcd是不是dcba的4倍就好了

#includevoidmain(){inti,j,k,m,n;printf("inputanumber:\n");scanf("%d",&n);i=n/1000;j=(n%1000)/100;k=(

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",

找人帮你做作业?再问：#include#defineNUM6voidmain(){intdata[NUM];inti,index,temp;for(i=0;i

#includevoidmain(){inta,b,c;scanf("%d%d",&a,&b);c=a;a=b;b=c;printf("%d%d\n",a,b);}

#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("%

#includeintis_four_digit(intvalue){return((value>=1000&&value=-9999&&value

//输入一个数字判断它是几位数并输出各位数字#includemain(){unsignedlongnum,i,count=0;printf("pleaseinputthenumber:");scanf

intmain(){inti,n;cin>>n;if(n

不明白楼主让用C语言,大家怎么都用C++ #include<stdio.h>void main(){ int num,temp,number=0,s

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