??

#include<stdio.h>
#define n 5
main()

{
int i=0;
int a;
char b[n];

char str;

while(str != '\n')
{
str=getchar();
a=i;
b[i]=str;
i++;

}

printf("%d\n",a);

for(i = 0;i < a;i++)
{
printf("%c",b[i]);
if(i != (a-1) )
printf("*");
}
printf("\n");
}

这样好理解 不过庸俗点

这样好理解.

[CODE]

#include <iostream.h>

int main()
{
int n,a,b,c,d,e;
cin>>n;
if(n<10)
cout<<n<<endl;
else if(n<100)
{
a = n/10;
b = n%10;
cout<<a<<"*"<<b<<endl;
} else if(n<1000)
{
a = n/100;
b = n/10;
b = b/10;
c = n%100;
cout<<a<<"*"<<b<<"*"<<c<<endl;
}else if(n<10000)
{
a = n/1000;
b = n/100;
b = b%10;
c = n%100;
c = c/10;
d = n%10;
cout<<a<<"*"<<b<<"*"<<c<<"*"<<d<<endl;
}else if(n<100000)
{
a = n/10000;
b = n/1000;
b = b%10;
c = n/100;
c = c%10;
d = n%1000;
d = n/10;
e = n%10000;
cout<<a<<"*"<<b<<"*"<<c<<"*"<<d<<"*"<<e<<endl;
}
return 0;
}

[/CODE]
最笨的方法，希望对你有帮助！！

#include<stdio.h>
#define N 80
main()
{
int i=0,j,k=0;
char c[N];
gets(c);
while(c[++i]!=0);
j=i;
printf("number:%d\n",j);
for(i=0;i<j;i++)
{(k==0)?printf("%c",c[i]):printf("*%c",c[i]);
k++;
}
getch();
return 0;
}

#include<stdio.h>
#define N 80
main()
{
int i=0,k=0;
char c[N];
gets(c);
while(c[++i]!=0);
printf("number:%d\n",i);
i=0;
while(c[i]!=0)
{(k==0)?printf("%c",c[i]):printf("*%c",c[i]);
k++;i++;
}
getch();
return 0;
}

[此贴子已经被作者于2006-11-9 15:54:45编辑过]

#include<stdio.h>
void main()
{
int i=0,j,num[5];
long a;
printf("请输入一个不多于5位的数：");
scanf("%d",&a);
do{
num[i]=a%10;
a/=10;
i++;
}while(a!=0);
printf("这是一个%d位数\n",i);
printf("%d",num[i-1]);
for(j=i-2;j>=0;j--)
printf("*%d",num[j]);
printf("\n");

}

我不知道为什么你们写那么多？？
是不是那样才算标准？？？看上面我的没错吧？？？

#include<stdio.h>
#define N 80
main()
{
int i;
char c[N];
gets(c);
for(i=0;c[i]!=0;i++)
(i==0)?printf("%c",c[i]):printf("*%c",c[i]);
printf("\nnumber:%d\n",i);
return 0;
}

[此贴子已经被作者于2006-11-9 17:30:18编辑过]

[CODE]#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int n = 0;
char *s = new char[6];
printf("Please input the number:\n");
scanf("%d",&n);

while(n>99999)
{
printf("This is a too big number!\n");
printf("Please input the number again:\n");
scanf("%d",&n);
}

itoa(n,s,10);

for(int i=0;i<(unsigned int)strlen(s);i++)
{
if(s[i+1]!=NULL)
{
printf("%c*",s[i]);

}else
printf("%c\n",s[i]);
}
delete []s;
return 0;
}[/CODE]

这种做法也可以！

[此贴子已经被作者于2006-11-10 15:09:02编辑过]

#include<stdio.h>
#define N 80
main()
{
int i;
char c[N];
gets(c);
for(i=0;c[i]!=0;i++)
(i==0)?printf("%c",c[i]):printf("*%c",c[i]);
printf("\nnumber:%d\n",i);
return 0;
}


有了这个其它人的都不用看了,语句最少,,执行时间最短,又没有位数限制,支持