【ICPC】The 2024 ICPC Kunming Invitational Contest E

Relearn through Review

#数论 #枚举 #gcd

题目描述

Given an integer sequence $a_1,a_2,\cdots ,a_n$ of length $n$ and a non-negative integer $k$, you can perform the following operation at most once: Choose two integers $l$ and $r$ such that $1\le l\le r\le n$, then for each $l\le i\le r$, change $a_i$ into $(a_i+k)$.

Maximize the greatest common divisor of the whole sequence.

An integer $g$ is said to be the common divisor of the whole sequence, if $a_i$ is divisible by $g$ for all $1\le i\le n$.

输入格式

There are multiple test cases. The first line of the input contains an integer $T$ indicating the number of test cases. For each test case:

The first line contains two integers $n$ and $k$ ($1\le n\le 3\times 10^5$, $0\le k\le 10^{18}$).

The second line contains $n$ integers $a_1,a_2,\cdots ,a_n$ ($1\le a_i\le 10^{18}$) indicating the sequence.

It’s guaranteed that the sum of $n$ of all test cases does not exceed $3\times 10^5$.

输出格式

For the first sample test case, choose $l=2$ and $r=4$. The sequence will become $\{5,5,15,10,10,555\}$. Its greatest common divisor is $5$.

样例 #1

样例输入 #1

2
6 2
5 3 13 8 10 555
3 0
3 6 9

样例输出 #1

5
3

解题思路

首先我们处理出$b$数组，其中$b_i:=a_i+k$，我们就能将问题转换为另一个问题：选择$b_i$中的区间$[l,r]$，使用$[b_l,b_r]$替换$a$中的$[a_l,a_r]$(可以不换)，求这样最大值。

显然，我们可以枚举右端点，由于前缀$gcd$只有$lo{g}_{2}n$种，因此我们可以使用$std::set$直接存入所有可能的交换组合，就能枚举出所有的情况了。

而对于后缀我们直接就预处理一个后缀$gcd$即，答案就是${\min }_{r=1}^{r\le n}(\min (\gcd (x,s{a}_{r+1})x\in set))$，

整体时间复杂度在$nlo{g}^{2}n$，包括$gcd$的$lo{g}_{2}n$。

代码

void solve() {
	int n, k;
	std::cin >> n >> k;
	std::vector<int>a(n + 1), pa(n + 2), sa(n + 2);
 
	for (int i = 1; i <= n; ++i) {
		std::cin >> a[i];
		pa[i] = std::gcd(pa[i - 1], a[i]);
	}
 
	std::vector<int>b(n + 1);
	for (int i = 1; i <= n; ++i) {
		b[i] = a[i] + k;
	}
 
	for (int i = n; i; --i) {
		sa[i] = std::gcd(sa[i + 1], a[i]);
	}
 
	int ans = pa[n];
 
	std::set<int>st[2];
	for (int r = 1; r <= n; ++r) {
		int now = r & 1, lst = now ^ 1;
 
		for (auto& x : st[lst]) {
			int t = std::gcd(x, b[r]);
			st[now].insert(t);
		}
 
		int t = std::gcd(pa[r - 1], b[r]);
		st[now].insert(t);
 
		st[lst].clear();
 
		for (auto& x : st[now]) {
			int w = std::gcd(x, sa[r + 1]);
			ans = std::max(ans, w);
		}
 
	}
 
	std::cout << ans << "\n";
}
 
 
signed main() {
	std::ios::sync_with_stdio(0);
	std::cin.tie(0);
	std::cout.tie(0);
 
	int t = 1;
	std::cin >> t;
 
	while (t--) {
		solve();
	}
}