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>

int dp[80001][11];//[i][mod]
int ans[11];
int main() {
	char str[80001];
	while (1) {
		for (int i = 0; i < 80001; i++) {
			for (int j = 0; j < 11; j++) {
				dp[i][j] = 0;
			}
		}
		for (int i = 0; i < 11; i++) {
			ans[i] = 0;
		}
		scanf("%s", str);
		if (strcmp(str, "0")==0)break;
		int length = strlen(str);
		dp[1][str[0] - '0']++;
		for (int i = 2; i <= length; i++) {
			int digit = str[i - 1] - '0';
			for (int mod = 0; mod <= 10; mod++) {
				dp[i][((mod * 10) + digit) % 11] = dp[i - 1][mod];
			}
			if(digit!=0)dp[i][digit]++;
			
			for (int mod = 0; mod <= 10; mod++) {
				ans[mod] += dp[i][mod];
			}
		}
		printf("%d\n", ans[0]);
	}
	return 0;
}