Algorithm

D. Good Sequences
time limit per test
2 seconds
memory limit per test
256 megabytes
problem link- https://codeforces.com/contest/265/problem/D
input
standard input
output
standard output

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a1, a2, ..., an are good.

Now she is interested in good sequences. A sequence x1, x2, ..., xk is called good if it satisfies the following three conditions:

  • The sequence is strictly increasing, i.e. xi < xi + 1 for each i (1 ≤ i ≤ k - 1).
  • No two adjacent elements are coprime, i.e. gcd(xi, xi + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q).
  • All elements of the sequence are good integers.

Find the length of the longest good sequence.

Input

The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 105) — the number of good integers. The second line contains a single-space separated list of good integers a1, a2, ..., an in strictly increasing order (1 ≤ ai ≤ 105; ai < ai + 1).

Output

Print a single integer — the length of the longest good sequence.

Examples
input
Copy
5
2 3 4 6 9
output
Copy
4
input
Copy
9
1 2 3 5 6 7 8 9 10
output
Copy
4
Note

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming


#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
int n, a[N], dp[N];
vector<vector<int> > g, d;

int rec(int idx) {
  if(idx == n)
  	return 0;

  int &ret = dp[idx];
  if(ret != -1)
  	return ret;
  ret = 0;

  for(int i = 0, cur; i < d[idx].size(); ++i) {
  	cur = upper_bound(g[d[idx][i]].begin(), g[d[idx][i]].end(), idx) - g[d[idx][i]].begin();
  	if(cur != g[d[idx][i]].size())
  		ret = max(ret, rec(g[d[idx][i]][cur]) + 1);
  }

  return ret;
}

int main() {
	g.resize(N);
	d.resize(N);

  scanf("%d", &n);
  for(int i = 0; i < n; ++i) {
    scanf("%d", a + i);

    int x = a[i], sqrtx = sqrt(x);
    for(int j = 2, fr; j <= sqrtx; ++j) {
    	fr = 0;
    	while(x % j == 0)
    		++fr, x /= j;
    	if(fr != 0)
    		g[j].push_back(i), d[i].push_back(j);
    }
    if(x != 1)
    	g[x].push_back(i), d[i].push_back(x);
  }

  memset(dp, -1, sizeof dp);
  int res = 0;
  for(int i = 0; i < n; ++i)
    res = max(res, rec(i) + 1);
  printf("%d\n", res);

  return 0;
}
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Input

x
–
+
cmd
5
2 3 4 6 9

Output

x
–
+
cmd
4
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Demonstration

Codeforces Solution-Good Sequences-Solution in C, C++, Java, Python

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