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 n n integers a 1 ,   a 2 ,   . . . ,   a n a_1, a_2, ..., a_n a1,a2,...,an are good.

Now she is interested in good sequences. A sequence x 1 ,   x 2 ,   . . . ,   x k x_1, x_2, ..., x_k x1,x2,...,xk is called good if it satisfies the following three conditions:

The sequence is strictly increasing, i.e. x i   &lt;   x i   +   1 x_i &lt; x_i + 1 xi<xi+1 for each i ( 1   ≤   i   ≤   k   −   1 ) i (1 ≤ i ≤ k - 1) i(1ik1).
No two adjacent elements are coprime, i.e. g c d ( x i ,   x i   +   1 )   &gt;   1 gcd(x_i, x_{i + 1}) &gt; 1 gcd(xi,xi+1)>1 for each i ( 1   ≤   i   ≤   k   −   1 ) i (1 ≤ i ≤ k - 1) i(1ik1) (where g c d ( p ,   q ) gcd(p, q) gcd(p,q) denotes the greatest common divisor of the integers p p p and q q 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   ≤   1 0 5 ) n (1 ≤ n ≤ 10^5) n(1n105) — the number of good integers. The second line contains a single-space separated list of good integers a 1 ,   a 2 ,   . . . ,   a n a_1, a_2, ..., a_n a1,a2,...,an in strictly increasing order ( 1   ≤   a i   ≤   1 0 5 ; a i   &lt;   a i   +   1 ) (1 ≤ a_i ≤ 10^5; a_i &lt; a_{i + 1}) (1ai105;ai<ai+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 ] [2; 4; 6; 9], [2; 4; 6], [3; 9], [6] [2;4;6;9],[2;4;6],[3;9],[6]. The length of the longest good sequence is 4 4 4.

Solution

DP
d p [ i ] dp[i] dp[i]表示以 i i i结尾的Good Sequence
d [ i ] d[i] d[i]表示所有含有 i i i这个质因子的 x x x m a x ( d p [ x ] ) max(dp[x]) max(dp[x])
每次转移完 d p [ x ] dp[x] dp[x]之后,再更新一下所有的 d [ i ] d[i] 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;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值