Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	static int[][] map;
	static int[][] directions8= {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
	static int[][] directions4= {{-1,0},{1,0},{0,-1},{0,1}};
	static long ans;
	static int[] parent;//union-find用
	static int[] rank;//union-find用

	public static void main(String[] args) throws IOException {

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int P = sc.nextInt();

		String S = sc.next();

		//今までに出たMod値の数
		long [] modP = new long[P];

		ans = 0;
		int nowMod = 0;
		int cardinal = 1;

		for(int digit = N-1;digit>=0 ; digit--) {
			int n = Character.getNumericValue(S.charAt(digit))*cardinal;
				nowMod +=n % P;
				nowMod%=P;
				if(nowMod==0) {
					ans++;
				}
				ans+=modP[nowMod];
				modP[nowMod]++;
				cardinal*=10;
				cardinal%=P;
		}

		System.out.println(ans);
	}

	//素数判定
	public static boolean isPrime(long num) {
		if (num < 2) return false;
		else if (num == 2) return true;
		else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

		double sqrtNum = Math.sqrt(num);
		for (int i = 3; i <= sqrtNum; i += 2)
		{
			if (num % i == 0)
			{
				// 素数ではない
				return false;
			}
		}

		// 素数である
		return true;
	}

	//BFS用に二つの配列を足し算する
	static int[] addArrayElms(int[] a, int[] b) {
		int[] c = new int[a.length];
		for(int i = 0; i < a.length; i++) {
			c[i] = a[i] + b[i];
		}
		return c;
	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
		return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
		return b == 0 ? a: gcd(b, a % b);
	}

	//Union-Find用の配列を要素数nで初期化
	static void ufInit(int n){
		parent = new int[n];
		rank = new int[n];
	}

	//Union-Findの要素の最上位の親（グループ名）を返す
	static int ufFind(int a) {
		if(parent[a]==a) {
			return a;
		}else {
			return ufFind(parent[a]);
		}
	}

	//Union-Find木を連結する
	static void ufUnite(int a, int b) {
		a = parent[a];
		b = parent[b];

		if(a==b) {
			return;
		}
		if(rank[a] > rank[b]){
			parent[b]=a;
		}else {
			parent[a]=b;
			if(rank[a]==rank[b]) {
				rank[b]++;
			}
		}
	}

	static boolean ufSame(int a, int b) {
		return ufFind(a)==ufFind(b);
	}
}