CF452F Permutation 与 P2757 [国家集训队] 等差子序列题解

合集 - codeforces题解集1(37)

1.CF1916E Happy Life in University 题解2023-12-31 2.CF763E Timofey and our friends animals题解2024-01-03 3.CF1270G Subset with Zero Sum2024-01-05 4.CF1045G AI robots题解2024-01-05 5.CF940F Machine Learning题解2024-01-06 6.CF678F Lena and Queries题解2024-01-12 7.CF1921F Sum of Progression 题解2024-01-16 8.CF526F Pudding Monsters 题解2024-01-21 9.CF455D Serega and Fun 题解2024-01-22 10.CF351D Jeff and Removing Periods 题解2024-01-22

11.CF452F Permutation 与 P2757 [国家集训队] 等差子序列题解2024-01-23

12.CF911G Mass Change Queries 题解2024-01-24 13.CF145E Lucky Queries 题解2024-01-25 14.CF1515F Phoenix and Earthquake 题解2024-01-26 15.CF765F Souvenirs 题解2024-01-27 16.CF1764H Doremy's Paint 2 题解2024-01-29 17.CF1000F One Occurrence题解2024-01-30 18.CF813E Army Creation 题解2024-01-31 19.CF1706E Qpwoeirut and Vertices 题解2024-02-01 20.CF620E New Year Tree 题解2024-02-02 21.CF1454F Array Partition 题解2024-02-03 22.CF1771F Hossam and Range Minimum Query 题解2024-02-08 23.CF1931F Chat Screenshots 另一种题解2024-02-14 24.CF896C Willem, Chtholly and Seniorious 题解2024-02-15 25.CF55D Beautiful numbers 题解2024-02-16 26.CF916E Jamie and Tree 题解2024-02-23 27.CF696B Puzzles 题解2024-02-28 28.CF383C Propagating tree 题解2024-02-29 29.CF1436E Complicated Computations 题解2024-03-08 30.CF817F MEX Queries 题解2024-03-20 31.CF1638E Colorful Operations 题解2024-03-21 32.CF1618G Trader Problem 题解2024-03-22 33.CF794F Leha and security system 题解2024-03-23 34.CF371E Subway Innovation 题解2024-04-04 35.CF1200E Compress Words 题解2024-04-06 36.CF1884D Counting Rhyme 题解2024-05-17 37.CF1982F Sorting Problem Again 题解2024-06-26

两道基本一样的题：

题目链接：

P2757 [国家集训队] 等差子序列

Permutation 链接：CF 或者洛谷

等差子序列那题其实就是长度不小于 $3$ 的等差数列是否存在，我们考虑等于 $3$ 的是否存在就行，因为等于 $3$ 长度的都不存在，更长的就不可能了，然后多了一个多测前提，就没啥了。两题其实基本一致。

考虑一下等差数列的三项，我们考虑中间那项设做 $m i d$ ，那么很显然一点，假如有公差为 $d i f f$ ，那么应该有 $m i d - d i f f$ 和 $m i d + d i f f$ 为它的前后两项，而这个 $d i f f$ 直接去确认它的存在性，过于困难，不仅要保证值的存在，值还应该处于当前 $m i d$ 所在值的两侧。而且枚举 $d i f f$ 过于慢，那么其实这个时候我们可以考虑 “正难则反” 思想。既然找存在性很困难，不如考虑不存在性的条件：

注意到原序列为排列，所以里面的数不重不漏且唯一。
基于第一点，那么对于 $m i d - d i f f$ 或者 $m i d + d i f f$ 要么在 $m i d$ 之前要么在它之后，我们如果从左往右更新到了 $m i d$ 发现没有找到 $m i d - d i f f$ 或者 $m i d + d i f f$ ，那么它们二者不存在于 $m i d$ 前面的一定在它后面的。

以这张图为例，比如从左往依次加入值，这里我们可以考虑把所有值放在一个轴上，类似权值线段树，在这个轴处记录加入这个值，考虑比如 $m i d - d i f f$ 很显然它有可能在 $m i d$ 之前更新，也有可能在之后更新，总而言之，它至少应该在某一侧。假如 $m i d - d i f f$ 与 $m i d + d i f f$ 都在同一侧，显然这种情况是不可取的了。现在我们考虑让 $d i f f$ 动起来，如果所有的 $d i f f$ 都不满足呢？很显然有这么一张图：

在权值轴上，这些数不同的 $d i f f$ 都出现了，因为我们都是从左往右更新的，所以一旦出现了说明一定是在 $m i d$ 的左侧下标范围出现的，都没出现就是都在右边下标范围，我们惊讶的发现，如果全都在同侧，那么很显然，在这个所有的 $d i f f$ 出现的数是对称？

具体的，我们假如出现过的数标 $1$ ，没出现的标 $0$ ，那么从左往右依次加入每个数，也依次枚举每个数当 $m i d$ ，我们惊讶的发现当前 $m i d$ 对应的权值轴情况如果是一个对称图形：

例如这样，就表示：所有的对应 $d i f f$ 对都是处于同一侧的，不存在异侧，即不存在解。。那问题就转变为了找不存在性即为判断权值轴是否是回文的，对称的，那自然而然的相当最简单的方法：维护权值轴信息自然而然用权值线段树，判断回文吗？考虑到线段树还可以同时维护信息，所以我们考虑使用哈希作为判断条件，为每个树节点维护两种哈希，一种是从左往右，一种从右往左，具体的：

例如： $1011$ 这个数从左往右就是 $1011$ 的哈希，从右往左就是 $1101$ 的哈希。如果两种哈希相同，自然就是回文对称了。

线段树维护哈希挺简单的，首先确定模数，然后确定下底数，这里就写单哈希了。具体的，再预处理个底数的幂次方数组，当合并线段树节点信息 $p u s h U p$ 的时候，注意是哪种哈希，从左往右的自然是，左边子树往右移动了 $r i g h t L e n$ 位，举个例子：

显然从左往右合并的答案为： $1011$ ，从右往左就是 $1110$ 了，谁拼谁罢了。乘个偏移信息以后加上另一边然后记得取模就行了。如果不满足回文，显然就是有解了，至此问题全部解决。

CF题目参考代码

 #include <bits/stdc++.h>
 
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)
 
// #define isPbdsFile
 
#ifdef isPbdsFile
 
#include <bits/extc++.h>
 
#else
 
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
 
#endif
 
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
 
template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}
 
template <typename T>
T lowBit(T x)
{
    return x & -x;
}
 
template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}
 
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}
 
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}
 
template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}
 
template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}
 
template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}
 
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}
 
 
template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;
 
    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }
 
    T3() { one = tow = three = 0; }
 
    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};
 
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}
 
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}
 
constexpr int N = 3e5 + 10;
constexpr int MX = 3e5;
constexpr int MOD = 1e9 + 7;
constexpr int BASE = 33;
ll powBase[N];
 
struct Node
{
    ll val, revVal; //正反哈希
    int len; //区间长度
} node[N << 2];
 
int n;
#define val(x) node[x].val
#define revVal(x) node[x].revVal
#define len(x) node[x].len
 
inline void push_up(const int curr)
{
    val(curr) = (val(ls(curr)) * powBase[len(rs(curr))] + val(rs(curr))) % MOD;
    revVal(curr) = (revVal(ls(curr)) + revVal(rs(curr)) * powBase[len(ls(curr))]) % MOD;
}
 
inline void add(const int curr, const int pos, const int l = 1, const int r = n)
{
    const int mid = l + r >> 1;
    if (l == r)
    {
        val(curr) = revVal(curr) = 1;
        return;
    }
    if (pos <= mid)add(ls(curr), pos, l, mid);
    else add(rs(curr), pos, mid + 1, r);
    push_up(curr);
}
 
inline pll query(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (l <= s and e <= r)return pll(val(curr),revVal(curr));
    const int mid = s + e >> 1;
    if (r <= mid)return query(ls(curr), l, r, s, mid);
    if (l > mid)return query(rs(curr), l, r, mid + 1, e);
    auto [leftVal,leftRevVal] = query(ls(curr), l, mid, s, mid);
    auto [rightVal,rightRevVal] = query(rs(curr), mid + 1, r, mid + 1, e);
    auto ansVal = (leftVal * powBase[r - mid] + rightVal) % MOD;
    auto ansRevVal = (leftRevVal + rightRevVal * powBase[mid - l + 1]) % MOD;
    return pll(ansVal, ansRevVal);
}
 
inline void build(const int curr = 1, const int l = 1, const int r = n)
{
    len(curr) = r - l + 1;
    val(curr) = revVal(curr) = 0;
    const int mid = l + r >> 1;
    if (l == r)return;
    build(ls(curr), l, mid);
    build(rs(curr), mid + 1, r);
}
 
int x;
 
inline void solve()
{
    cin >> n;
    build();
    bool check = false;
    forn(i, 1, n)
    {
        cin >> x;
        auto minLen = min(x - 1, n - x);
        auto [val1,val2] = query(1, x - minLen, x + minLen);
        if (val1 != val2)check = true;
        add(1, x);
    }
    cout << (check ? "YES" : "NO") << endl;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    powBase[0] = 1;
    forn(i, 1, MX)powBase[i] = powBase[i - 1] * BASE % MOD;
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

洛谷题目参考代码

 #include <bits/stdc++.h>
 
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)
 
// #define isPbdsFile
 
#ifdef isPbdsFile
 
#include <bits/extc++.h>
 
#else
 
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
 
#endif
 
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
 
template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}
 
template <typename T>
T lowBit(T x)
{
    return x & -x;
}
 
template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}
 
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}
 
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}
 
template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}
 
template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}
 
template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}
 
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}
 
 
template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;
 
    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }
 
    T3() { one = tow = three = 0; }
 
    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};
 
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}
 
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}
 
constexpr int N = 5e5 + 10;
constexpr int MX = 5e5;
constexpr int MOD = 1e9 + 7;
constexpr int BASE = 33;
ll powBase[N];
 
struct Node
{
    ll val, revVal; //正反哈希
    int len; //区间长度
} node[N << 2];
 
int n;
#define val(x) node[x].val
#define revVal(x) node[x].revVal
#define len(x) node[x].len
 
inline void push_up(const int curr)
{
    val(curr) = (val(ls(curr)) * powBase[len(rs(curr))] + val(rs(curr))) % MOD;
    revVal(curr) = (revVal(ls(curr)) + revVal(rs(curr)) * powBase[len(ls(curr))]) % MOD;
}
 
inline void add(const int curr, const int pos, const int l = 1, const int r = n)
{
    const int mid = l + r >> 1;
    if (l == r)
    {
        val(curr) = revVal(curr) = 1;
        return;
    }
    if (pos <= mid)add(ls(curr), pos, l, mid);
    else add(rs(curr), pos, mid + 1, r);
    push_up(curr);
}
 
inline pll query(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (l <= s and e <= r)return pll(val(curr),revVal(curr));
    const int mid = s + e >> 1;
    if (r <= mid)return query(ls(curr), l, r, s, mid);
    if (l > mid)return query(rs(curr), l, r, mid + 1, e);
    auto [leftVal,leftRevVal] = query(ls(curr), l, mid, s, mid);
    auto [rightVal,rightRevVal] = query(rs(curr), mid + 1, r, mid + 1, e);
    auto ansVal = (leftVal * powBase[r - mid] + rightVal) % MOD;
    auto ansRevVal = (leftRevVal + rightRevVal * powBase[mid - l + 1]) % MOD;
    return pll(ansVal, ansRevVal);
}
 
inline void build(const int curr = 1, const int l = 1, const int r = n)
{
    len(curr) = r - l + 1;
    val(curr) = revVal(curr) = 0;
    const int mid = l + r >> 1;
    if (l == r)return;
    build(ls(curr), l, mid);
    build(rs(curr), mid + 1, r);
}
 
int x;
 
inline void solve()
{
    cin >> n;
    build();
    bool check = false;
    forn(i, 1, n)
    {
        cin >> x;
        auto minLen = min(x - 1, n - x);
        auto [val1,val2] = query(1, x - minLen, x + minLen);
        if (val1 != val2)check = true;
        add(1, x);
    }
    cout << (check ? "Y" : "N") << endl;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    powBase[0] = 1;
    forn(i, 1, MX)powBase[i] = powBase[i - 1] * BASE % MOD;
    int test = 1;
    //    read(test);
    cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

单 次 T e s t 的 时 间 复 杂 度 为 O (n \log n)

posted @ 2024-01-23 20:54 Athanasy 阅读(19) 评论(0) 编辑收藏举报

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Athanasy

CF452F Permutation 与 P2757 [国家集训队] 等差子序列题解

两道基本一样的题：

题目链接：

Permutation 链接：CF 或者洛谷

等差子序列那题其实就是长度不小于 $3$ 的等差数列是否存在，我们考虑等于 $3$ 的是否存在就行，因为等于 $3$ 长度的都不存在，更长的就不可能了，然后多了一个多测前提，就没啥了。两题其实基本一致。

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	#include <bits/stdc++.h>

	// #pragma GCC optimize("Ofast,unroll-loops")
	// #pragma GCC optimize(2)

	// #define isPbdsFile

	#ifdef isPbdsFile

	#include <bits/extc++.h>

	#else

	#include <ext/pb_ds/priority_queue.hpp>
	#include <ext/pb_ds/hash_policy.hpp>
	#include <ext/pb_ds/tree_policy.hpp>
	#include <ext/pb_ds/trie_policy.hpp>
	#include <ext/pb_ds/tag_and_trait.hpp>
	#include <ext/pb_ds/hash_policy.hpp>
	#include <ext/pb_ds/list_update_policy.hpp>
	#include <ext/pb_ds/assoc_container.hpp>
	#include <ext/pb_ds/exception.hpp>
	#include <ext/rope>

	#endif

	using namespace std;
	using namespace __gnu_cxx;
	using namespace __gnu_pbds;
	typedef long long ll;
	typedef long double ld;
	typedef pair<int, int> pii;
	typedef pair<ll, ll> pll;
	typedef tuple<int, int, int> tii;
	typedef tuple<ll, ll, ll> tll;
	typedef unsigned int ui;
	typedef unsigned long long ull;
	typedef __int128 i128;
	#define hash1 unordered_map
	#define hash2 gp_hash_table
	#define hash3 cc_hash_table
	#define stdHeap std::priority_queue
	#define pbdsHeap __gnu_pbds::priority_queue
	#define sortArr(a, n) sort(a+1,a+n+1)
	#define all(v) v.begin(),v.end()
	#define yes cout<<"YES"
	#define no cout<<"NO"
	#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
	#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
	#define forn(i, a, b) for(int i = a; i <= b; i++)
	#define forv(i, a, b) for(int i=a;i>=b;i--)
	#define ls(x) (x<<1)
	#define rs(x) (x<<1\|1)
	#define endl '\n'
	//用于Miller-Rabin
	[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

	template <typename T>
	int disc(T* a, int n)
	{
	return unique(a + 1, a + n + 1) - (a + 1);
	}

	template <typename T>
	T lowBit(T x)
	{
	return x & -x;
	}

	template <typename T>
	T Rand(T l, T r)
	{
	static mt19937 Rand(time(nullptr));
	uniform_int_distribution<T> dis(l, r);
	return dis(Rand);
	}

	template <typename T1, typename T2>
	T1 modt(T1 a, T2 b)
	{
	return (a % b + b) % b;
	}

	template <typename T1, typename T2, typename T3>
	T1 qPow(T1 a, T2 b, T3 c)
	{
	a %= c;
	T1 ans = 1;
	for (; b; b >>= 1, (a = a) %= c)if (b & 1)(ans = a) %= c;
	return modt(ans, c);
	}

	template <typename T>
	void read(T& x)
	{
	x = 0;
	T sign = 1;
	char ch = getchar();
	while (!isdigit(ch))
	{
	if (ch == '-')sign = -1;
	ch = getchar();
	}
	while (isdigit(ch))
	{
	x = (x << 3) + (x << 1) + (ch ^ 48);
	ch = getchar();
	}
	x *= sign;
	}

	template <typename T, typename... U>
	void read(T& x, U&... y)
	{
	read(x);
	read(y...);
	}

	template <typename T>
	void write(T x)
	{
	if (typeid(x) == typeid(char))return;
	if (x < 0)x = -x, putchar('-');
	if (x > 9)write(x / 10);
	putchar(x % 10 ^ 48);
	}

	template <typename C, typename T, typename... U>
	void write(C c, T x, U... y)
	{
	write(x), putchar(c);
	write(c, y...);
	}


	template <typename T11, typename T22, typename T33>
	struct T3
	{
	T11 one;
	T22 tow;
	T33 three;

	bool operator<(const T3 other) const
	{
	if (one == other.one)
	{
	if (tow == other.tow)return three < other.three;
	return tow < other.tow;
	}
	return one < other.one;
	}

	T3() { one = tow = three = 0; }

	T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
	{
	}
	};

	template <typename T1, typename T2>
	void uMax(T1& x, T2 y)
	{
	if (x < y)x = y;
	}

	template <typename T1, typename T2>
	void uMin(T1& x, T2 y)
	{
	if (x > y)x = y;
	}

	constexpr int N = 3e5 + 10;
	constexpr int MX = 3e5;
	constexpr int MOD = 1e9 + 7;
	constexpr int BASE = 33;
	ll powBase[N];

	struct Node
	{
	ll val, revVal; //正反哈希
	int len; //区间长度
	} node[N << 2];

	int n;
	#define val(x) node[x].val
	#define revVal(x) node[x].revVal
	#define len(x) node[x].len

	inline void push_up(const int curr)
	{
	val(curr) = (val(ls(curr)) * powBase[len(rs(curr))] + val(rs(curr))) % MOD;
	revVal(curr) = (revVal(ls(curr)) + revVal(rs(curr)) * powBase[len(ls(curr))]) % MOD;
	}

	inline void add(const int curr, const int pos, const int l = 1, const int r = n)
	{
	const int mid = l + r >> 1;
	if (l == r)
	{
	val(curr) = revVal(curr) = 1;
	return;
	}
	if (pos <= mid)add(ls(curr), pos, l, mid);
	else add(rs(curr), pos, mid + 1, r);
	push_up(curr);
	}

	inline pll query(const int curr, const int l, const int r, const int s = 1, const int e = n)
	{
	if (l <= s and e <= r)return pll(val(curr),revVal(curr));
	const int mid = s + e >> 1;
	if (r <= mid)return query(ls(curr), l, r, s, mid);
	if (l > mid)return query(rs(curr), l, r, mid + 1, e);
	auto [leftVal,leftRevVal] = query(ls(curr), l, mid, s, mid);
	auto [rightVal,rightRevVal] = query(rs(curr), mid + 1, r, mid + 1, e);
	auto ansVal = (leftVal * powBase[r - mid] + rightVal) % MOD;
	auto ansRevVal = (leftRevVal + rightRevVal * powBase[mid - l + 1]) % MOD;
	return pll(ansVal, ansRevVal);
	}

	inline void build(const int curr = 1, const int l = 1, const int r = n)
	{
	len(curr) = r - l + 1;
	val(curr) = revVal(curr) = 0;
	const int mid = l + r >> 1;
	if (l == r)return;
	build(ls(curr), l, mid);
	build(rs(curr), mid + 1, r);
	}

	int x;

	inline void solve()
	{
	cin >> n;
	build();
	bool check = false;
	forn(i, 1, n)
	{
	cin >> x;
	auto minLen = min(x - 1, n - x);
	auto [val1,val2] = query(1, x - minLen, x + minLen);
	if (val1 != val2)check = true;
	add(1, x);
	}
	cout << (check ? "YES" : "NO") << endl;
	}

	signed int main()
	{
	// MyFile
	Spider
	//------------------------------------------------------
	// clock_t start = clock();
	powBase[0] = 1;
	forn(i, 1, MX)powBase[i] = powBase[i - 1] * BASE % MOD;
	int test = 1;
	// read(test);
	// cin >> test;
	forn(i, 1, test)solve();
	// while (cin >> n, n)solve();
	// while (cin >> test)solve();
	// clock_t end = clock();
	// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
	}

Athanasy

CF452F Permutation 与 P2757 [国家集训队] 等差子序列 题解

两道基本一样的题：

题目链接：

Permutation 链接：CF 或者 洛谷

等差子序列那题其实就是长度不小于 3 的等差数列是否存在，我们考虑等于 3 的是否存在就行，因为等于 3 长度的都不存在，更长的就不可能了，然后多了一个多测前提，就没啥了。两题其实基本一致。

公告

搜索

常用链接

合集

随笔分类

随笔档案

阅读排行榜

评论排行榜

推荐排行榜

最新评论

CF452F Permutation 与 P2757 [国家集训队] 等差子序列题解

Permutation 链接：CF 或者洛谷

等差子序列那题其实就是长度不小于 $3$ 的等差数列是否存在，我们考虑等于 $3$ 的是否存在就行，因为等于 $3$ 长度的都不存在，更长的就不可能了，然后多了一个多测前提，就没啥了。两题其实基本一致。