CodeForces 264B.Good Sequences(DP)

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博客围绕Good Sequences问题展开，该问题要求找出满足严格递增、相邻元素不互质且元素为给定好整数的最长序列长度。给出了输入输出示例，采用动态规划（DP）方法求解，定义了dp[i]和d[i]，每次转移完dp[x]后更新d[i]。

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 $i (1 \leq i \leq k - 1)$ .
No two adjacent elements are coprime, i.e. $gcd(x_i, x_{i + 1}) > 1$ for each $i (1 \leq i \leq k - 1)$ (where $g c d (p, q)$ 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

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
$d p [i]$ 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 $m a x (d p [x])$
每次转移完 $d p [x]$ 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;
}