description: Kuzya started going to school. He was given math homework in which he was given an array a of length n and an array of symbols b of length n, consisting of symbols '*' and '/'.
  
  Let's denote a path of calculations for a segment [l; r] (1 ≤ l ≤ r ≤ n) in the following way: 
  
    * Let x=1 initially. For every i from l to r we will consequently do the following: if b_i= '*', x=x*a_i, and if b_i= '/', then x=(x)/(a_i). Let's call a path of calculations for the segment [l; r] a list of all x that we got during the calculations (the number of them is exactly r - l + 1). 
  
  
  
  For example, let a=[7, 12, 3, 5, 4, 10, 9], b=[/, *, /, /, /, *, *], l=2, r=6, then the path of calculations for that segment is [12, 4, 0.8, 0.2, 2].
  
  Let's call a segment [l;r] simple if the path of calculations for it contains only integer numbers. 
  
  Kuzya needs to find the number of simple segments [l;r] (1 ≤ l ≤ r ≤ n). Since he obviously has no time and no interest to do the calculations for each option, he asked you to write a program to get to find that number!
  
  Input
  
  The first line contains a single integer n (2 ≤ n ≤ 10^6).
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).
  
  The third line contains n symbols without spaces between them — the array b_1, b_2 … b_n (b_i= '/' or b_i= '*' for every 1 ≤ i ≤ n).
  
  Output
  
  Print a single integer — the number of simple segments [l;r].
  
  Examples
  
  Input
  
  
  3
  1 2 3
  */*
  
  
  Output
  
  
  2
  
  
  Input
  
  
  7
  6 4 10 1 2 15 1
  */*/*//
  
  
  Output
  
  
  8
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 1e6 + 5;
  int a[N];
  char s[N];
  unordered_map<int, int> d[N];
  stack<unordered_map<int, int> *> q;
  vector<int> prime;
  int vis[1005];
  void euler() {
    for (int i = 2; i <= 1000; ++i) {
      if (!vis[i]) prime.push_back(i);
      for (int j = 0; j < prime.size() && i * prime[j] < 1000; ++j) {
        vis[i * prime[j]] = 1;
        if (i % prime[j] == 0) break;
      }
    }
  }
  void init() {
    for (int i = 1; i <= 1000000; ++i) {
      int x = i;
      for (int j : prime) {
        if (j * j > x) break;
        while (x % j == 0) {
          ++d[i][j];
          x /= j;
        }
      }
      if (x > 1) ++d[i][x];
    }
  }
  bool check(unordered_map<int, int> &x, unordered_map<int, int> &y) {
    for (auto &i : y) {
      if (x.find(i.first) == x.end() || x[i.first] < i.second) return false;
    }
    return true;
  }
  void merge(unordered_map<int, int> &x, unordered_map<int, int> &y) {
    for (auto &i : y) {
      x[i.first] += i.second;
    }
  }
  void sub(unordered_map<int, int> &x, unordered_map<int, int> &y) {
    for (auto &i : y) {
      x[i.first] -= i.second;
      if (!x[i.first]) x.erase(i.first);
    }
  }
  int main() {
    euler();
    init();
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    scanf("%s", s);
    long long ans = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i] == '*') {
        auto *mp = new unordered_map<int, int>;
        merge(*mp, d[a[i]]);
        q.push(mp);
      } else {
        auto *mp = new unordered_map<int, int>;
        merge(*mp, d[1]);
        q.push(mp);
        unordered_map<int, int> &cur = d[a[i]];
        while (!q.empty() && !check(*q.top(), cur)) {
          unordered_map<int, int> *p = q.top();
          q.pop();
          if (!q.empty()) {
            merge(*p, *q.top());
            delete q.top();
            q.pop();
            q.push(p);
          }
        }
        if (!q.empty()) sub(*q.top(), cur);
      }
      ans += q.size();
    }
    printf("%lld\n", ans);
    return 0;
  }
  