前言
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
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/282482.html