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 maxn = 1000005;
  int n, top, mx;
  long long ans;
  int a[maxn], stk[maxn];
  vector<int> p[maxn], v[maxn];
  string s;
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), mx = max(mx, a[i]);
    for (int i = 2; i <= mx; i++)
      if (v[i].empty())
        for (int j = i; j <= mx; j += i) {
          int k = j;
          while (k % i == 0) k /= i, v[j].push_back(i);
        }
    cin >> s, s = " " + s;
    for (int i = 1; i <= n; i++) {
      stk[++top] = i;
      if (s[i] == '*')
        for (int j = 0; j < v[a[i]].size(); j++) p[v[a[i]][j]].push_back(i);
      if (s[i] == '/')
        for (int j = 0; j < v[a[i]].size(); j++) {
          if (p[v[a[i]][j]].size() == 0) {
            top = 0;
            break;
          }
          while (top && stk[top] > p[v[a[i]][j]].back()) top--;
          p[v[a[i]][j]].pop_back();
        }
      ans += top;
    }
    printf("%lld\n", ans);
    return 0;
  }
  