Problem:
Problem H: Eleven Lover

Edward Leven loves multiples of eleven very much. When he sees a number, he always tries
to find consecutive subsequences (or substrings) forming multiples of eleven. He calls such
subsequences as 11-sequences. For example, he can find an 11-sequence 781 in a number 17819.

He thinks a number which has many 11-sequences is a good number. He would like to find out
a very good number. As the first step, he wants an easy way to count how many 11-sequences
are there in a given number. Even for him, counting them from a big number is not easy.
Fortunately, one of his friends, you, is a brilliant programmer. He asks you to write a program
to count the number of 11-sequences. Note that an 11-sequence must be a positive number
without leading zeros.

Input

The input is a sequence of lines each of which contains a number consisting of less than or equal
to 80000 digits.

The end of the input is indicated by a line containing a single zero, which should not be processed.

Output

For each input number, output a line containing the number of 11-sequences.

You can assume the answer fits in a 32-bit signed integer.

Sample Input

17819
1111
11011
1234567891011121314151617181920
0

Output for the Sample Input

1
4
4
38

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
char S[80010];
int i,j,dp[80010][20],t,r,l;
int main()
{
	for(;scanf("%s",S),S[0]!='0';)
	{
		memset(dp,0,sizeof(dp));
		l=strlen(S);
		for(i=r=0;i<l;i++)
		{
			t=S[i]-'0';
			dp[i+1][t]=!!t;
			for(j=0;j<11;j++)
				if(dp[i][j])
					dp[i+1][(10*j+t)%11]+=dp[i][j];
			r+=dp[i+1][0];
		}
		printf("%d\n",r);
	}
	return 0;
}