CF1548B 题解

前言

题目传送门！

更好的阅读体验？

做法：ST 表加尺取。

思路

看到同余，立刻想到作差。我们建立差分数组 /(c_i = |a_i – a_{i-1}|/)，注意取了绝对值。

此时，我们只需在 /(c_i/) 中寻找最长区间 /(/left[l, r/right]/)，使得 /(/gcd(c_l, c_{l+1}, /cdots, c_r) > 1/)。

这东西显然能用 ST 表维护。代码是模板，不需解释。

typedef long long LL;
LL gcd(LL x, LL y) {return y == 0 ? x : gcd(y, x%y);}
struct ST
{
	LL dp[N][20];
	void build()
	{
		for (int i = 1; i <= n; i++) dp[i][0] = c[i]; //c 数组是差分数组
		int k = log2(n);
		for (int j = 1; j <= k; j++)
			for (int i = 1; i + (1 << j) - 1 <= n; i++)
				dp[i][j] = gcd(dp[i][j-1], dp[i + (1 << (j-1))][j - 1]);
	}
	LL query(int l, int r)
	{
		int k = log2(r - l + 1);
		return gcd(dp[l][k], dp[r - (1 << k) + 1][k]);
	}
	
}st;

然后就是简单的尺取了，依次枚举右端点并更新左端点。

bool chk(int l, int r) {return st.query(l, r) > 1;}
int zlttql() //尺取，顺便膜拜神仙 zlt
{
	int maxn = 0;
	for (int l = 1, r = 1; r <= n; r++)
	{
		for (; l < r && !chk(l+1, r); l++);
		maxn = max(maxn, r - l + 1);
	}
	return maxn;
}

最后输出答案就完成了。

完整代码

#include <iostream>
#include <cstdio>
#include <cmath>
#define endl putchar('/n')
#define space putchar(' ')
using namespace std;
typedef long long LL;
LL read()
{
	char op = getchar(); LL x = 0, f = 1;
	while (op < 48 || op > 57) {if (op == '-') f = -1; op = getchar();}
	while (48 <= op && op <= 57) x = (x << 1) + (x << 3) + (op ^ 48), op = getchar();
	return x * f;
}
void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + 48);
}
const int N = 2e5 + 5;
LL a[N], c[N]; //c:差分 
int n;
LL gcd(LL x, LL y) {return y == 0 ? x : gcd(y, x%y);}
struct ST
{
	LL dp[N][20];
	void build()
	{
		for (int i = 1; i <= n; i++) dp[i][0] = c[i];
		int k = log2(n);
		for (int j = 1; j <= k; j++)
			for (int i = 1; i + (1 << j) - 1 <= n; i++)
				dp[i][j] = gcd(dp[i][j-1], dp[i + (1 << (j-1))][j - 1]);
	}
	LL query(int l, int r)
	{
		int k = log2(r - l + 1);
		return gcd(dp[l][k], dp[r - (1 << k) + 1][k]);
	}
	
}st;
bool chk(int l, int r) {return st.query(l, r) > 1;}
int zlttql() //尺取 
{
	int maxn = 0;
	for (int l = 1, r = 1; r <= n; r++)
	{
		for (; l < r && !chk(l+1, r); l++);
		maxn = max(maxn, r - l + 1);
	}
	return maxn;
}
int main()
{
    int T = read();
    while (T--)
    {
     	n = read();
        for (int i = 1; i <= n; i++) a[i] = read(), c[i] = abs(a[i] - a[i-1]);
        st.build();
        //差分后，需满足：一段区间 gcd > 1
        write(zlttql()), endl;
    }
	return 0;
}

希望能帮助到大家！
首发：2022-08-14 17:22:57