Count of Numbers in a Range divisible by m and having digit d in even positions

#include <bits/stdc++.h>

using namespace std;

#define N 100

void pre_process(bool dp[N][N], string s)

{

int n = s.size();

for (int i = 0; i < n; i++) {

for (int j = 0; j < n; j++)

dp[i][j] = false;

}

for (int j = 1; j <= n; j++) {

for (int i = 0; i <= n - j; i++) {

if (j <= 2) {

if (s[i] == s[i + j - 1])

dp[i][i + j - 1] = true;

}

else if (s[i] == s[i + j - 1])

dp[i][i + j - 1] = dp[i + 1][i + j - 2];

}

}

}

int countPairs(string s)

{

bool dp[N][N];

pre_process(dp, s);

int n = s.length();

int left[n];

memset(left, 0, sizeof left);

int right[n];

memset(right, 0, sizeof right);

left[0] = 1;

for (int i = 1; i < n; i++) {

for (int j = 0; j <= i; j++) {

if (dp[j][i] == 1)

left[i]++;

}

}

right[n - 1] = 1;

for (int i = n - 2; i >= 0; i--) {

right[i] = right[i + 1];

for (int j = n - 1; j >= i; j--) {

if (dp[i][j] == 1)

right[i]++;

}

}

int ans = 0;

for (int i = 0; i < n - 1; i++)

ans += left[i] * right[i + 1];

return ans;

}

int main()

{

string s = "abacaba";

cout << countPairs(s);

return 0;

}