CodeForces 264B.Good Sequences(DP)

B. Good Sequences

(2s,256M)
Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks $n$ integers $a_1, a_2, ..., a_n$ are good.

Now she is interested in good sequences. A sequence $x_1, x_2, ..., x_k$ is called good if it satisfies the following three conditions:

The sequence is strictly increasing, i.e. $x_i < x_i + 1$ for each $i (1 ≤ i ≤ k - 1)$ .
No two adjacent elements are coprime, i.e. $gcd(x_i, x_{i + 1}) > 1$ 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 $n (1 ≤ n ≤ 10^5)$ — the number of good integers. The second line contains a single-space separated list of good integers $a_1, a_2, ..., a_n$ in strictly increasing order $(1 ≤ a_i ≤ 10^5; a_i < a_{i + 1})$ .

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

Examples
input

5
2 3 4 6 9

output

4

input

9
1 2 3 5 6 7 8 9 10

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output

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 $4$ .

Solution

DP
$dp[i]$ 表示以 $i$ 结尾的Good Sequence
$d[i]$ 表示所有含有 $i$ 这个质因子的 $x$ 中 $max(dp[x])$
每次转移完 $dp[x]$ 之后，再更新一下所有的 $d[i]$ 即可

Code

#include <cstdio>
#include <algorithm>
#define N 100010

using namespace std;

int a[N], dp[N], d[N];
int gcd(int x, int y) {
    if (y == 0) return x;
    else return(y, x % y);
}

int main() {
    int n;
    scanf("%d", &n);
    if (n == 1) {
        printf("%d\n", 1);
        return 0;
    }
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        int e = a[i];
        for (int j = 2; j * j <= a[i]; ++j) {
            if (e % j == 0) {
                dp[a[i]] = max(dp[a[i]], d[j]);
                while (e % j == 0) e /= j;
            }
        }
        if (e > 1) dp[a[i]] = max(dp[a[i]], d[e]);
        dp[a[i]]++;
        ans = max(ans, dp[a[i]]);
        e = a[i];
        for (int j = 2; j * j <= a[i]; ++j) {
            if (e % j == 0) {
                d[j] = dp[a[i]];
                while (e % j == 0) e /= j;
            }
        }
        if (e > 1) d[e] = dp[a[i]];
    }
    printf("%d\n", ans);
    return 0;
}