最原始的方法 因为我只学到这

解题思路: 最原始的方法 肯定都看得懂

注意事项:

参考代码:#include<stdio.h>
int main(){
 int a;
 scanf("%d",&a);
 if(1<=(a/10000)&&(a/10000)<10)

{
 printf("5\n");

}
 else if(1<=(a/1000)&&(a/1000)<10)

{
 printf("4\n");

}
 else if(1<=(a/100)&&(a/100)<10)

{
 printf("3\n");

}
 else if(1<=(a/10)&&(a/10)<10)

{
 printf("2\n");

}
 else 

{
 printf("1\n");

}
 int b=a/10000;
 int c=(a-(b*10000))/1000;
 int d=(a-(b*10000)-(c*1000))/100;
 int e=(a-(b*10000)-(c*1000)-(d*100))/10;
 int f=(a-(b*10000)-(c*1000)-(d*100)-(e*10));
 if(1<=(a/10000)&&(a/10000)<10)

{
 printf("%d %d %d %d %d\n",b,c,d,e,f);

}
 else if(1<=(a/1000)&&(a/1000)<10)

{
 printf("%d %d %d %d\n",c,d,e,f);

}
 else if(1<=(a/100)&&(a/100)<10)

{
 printf("%d %d %d\n",d,e,f);

}
 else if(1<=(a/10)&&(a/10)<10)

{
 printf("%d %d\n",e,f);

}
 else

{
 printf("%d\n",f);

}
 if(1<=(a/10000)&&(a/10000)<10)

{
 printf("%d%d%d%d%d",f,e,d,c,b);

}
 else if(1<=(a/1000)&&(a/1000)<10)

{
 printf("%d%d%d%d",f,e,d,c);

}
 else if(1<=(a/100)&&(a/100)<10)

{
 printf("%d%d%d",f,e,d);

}
 else if(1<=(a/10)&&(a/10)<10)

{
 printf("%d%d",f,e);

}
 else

{
 printf("%d",f);

}
 return 0;
 
}