#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
void main()
{
	char s[100],str[100],*end,*c;
	long a,b,sum,n=1,k;int i;
	while(scanf("%s",str)&&str[0]!='0')
	{
		for(i=1;i<=10;i++)
		{
			k=strlen(str);
		    sum=strtol(str,&end,10);
			sum=sum+i*pow(10,k);
		if(sum%11==0)
		{b=sum/11;printf("%ld. %ld\n",n,b);break;}
		else if(i>=10)printf("%ld. IMPOSSIBLE\n",n);
		}
		n++;
	}
}

• Re:例子都对，就是过不了，running error！ (498) trueshlqsh 2008-03-07 20:03:18
• Each case is specified on a separate line and is made of a single positive number having less than 1,000,000 digits. (118) orangelegend 2008-04-14 21:31:14