c语言中输入一个正整数,逆序输出的编程

#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i

#include <stdio.h>void main(){    int n;   &

#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);

#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}

#includeintmain(){intn;charch;inti;printf("输入一个10以内的正整数和相应个数的五进制成绩：\n");scanf("%d",&n);for(i=0

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

#include <stdio.h>void main (){ int a[10]; int i,k=0,n,min,

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

#includeintmain(void){longintx;inty,i=1;printf("请输入一个不超过100000的整数:");//99999||x

#includeintmat[10][10];voidmain(){intn,i,j;intok=1;scanf("%d",&n);for(i=0;i

#includeintmain()#definen5{inta[n],m,b[n+1],temp,k,i;printf("n=%d\ntheorigalnumber:",n);for(i=0;i

#includevoidmain(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=

使用了数组的动态分配,看不懂的话我还有其他方案#include#includeboolprime(intnum){boolflag=true;if(num

x%10=a1,x/10%10=a2,x/100=a3,a=a1*100+a2*10+a3再答：判断是几位:if(a>100)3if(a>1000)4if(a>10000)5再问：再问：能帮我看一下错

#includeintmain(void){intn=0,m=0,i=0,j=0,k=0;scanf("%d",&n);while(n--){\x09scanf("%d",&m);\x09for(j=

#includevoidmain(){inti,n;printf("pleaseinputn:");scanf("%d",&n);printf("%d=",n);for(i=2;i

修改如下：1.主函数中intcountdigit,digit=2,number中,countdigit重复定义,改为其他变量,比如x2.改为scanf("%d",&number);3.printf("

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

#include#defineNUM20voidmain(){longn;intdata[NUM]={0},len=0;scanf("%d",&n);do{data[len++]=n%10;n/=10

#include"stdio.h"#include"math.h"intmain(void){intcount,i,m,n,sum;intrepeat,ri;intprime(intm);scanf(