CF1436E Complicated Computations 题解

合集 - codeforces题解集1(37)

1.CF1916E Happy Life in University 题解2023-12-312.CF763E Timofey and our friends animals题解2024-01-033.CF1270G Subset with Zero Sum2024-01-054.CF1045G AI robots题解2024-01-055.CF940F Machine Learning题解2024-01-066.CF678F Lena and Queries题解2024-01-127.CF1921F Sum of Progression 题解2024-01-168.CF526F Pudding Monsters 题解2024-01-219.CF455D Serega and Fun 题解2024-01-2210.CF351D Jeff and Removing Periods 题解2024-01-2211.CF452F Permutation 与 P2757 [国家集训队] 等差子序列 题解2024-01-2312.CF911G Mass Change Queries 题解2024-01-2413.CF145E Lucky Queries 题解2024-01-2514.CF1515F Phoenix and Earthquake 题解2024-01-2615.CF765F Souvenirs 题解2024-01-2716.CF1764H Doremy's Paint 2 题解2024-01-2917.CF1000F One Occurrence题解2024-01-3018.CF813E Army Creation 题解2024-01-3119.CF1706E Qpwoeirut and Vertices 题解2024-02-0120.CF620E New Year Tree 题解2024-02-0221.CF1454F Array Partition 题解2024-02-0322.CF1771F Hossam and Range Minimum Query 题解2024-02-0823.CF1931F Chat Screenshots 另一种题解2024-02-1424.CF896C Willem, Chtholly and Seniorious 题解2024-02-1525.CF55D Beautiful numbers 题解2024-02-1626.CF916E Jamie and Tree 题解2024-02-2327.CF696B Puzzles 题解2024-02-2828.CF383C Propagating tree 题解2024-02-29

29.CF1436E Complicated Computations 题解2024-03-08

30.CF817F MEX Queries 题解2024-03-2031.CF1638E Colorful Operations 题解2024-03-2132.CF1618G Trader Problem 题解2024-03-2233.CF794F Leha and security system 题解2024-03-2334.CF371E Subway Innovation 题解2024-04-0435.CF1200E Compress Words 题解2024-04-0636.CF1884D Counting Rhyme 题解2024-05-1737.CF1982F Sorting Problem Again 题解2024-06-26

收起

题目链接：CF 或者 洛谷

关于 $mex$ 问题是一个比较久远的问题，有很多经典的方法去解决。本题的 $mex$ 是从正整数开始的，不要忽略掉。

来讲讲常见的两种解决方案，首先回到题目所问，如果我们暴力地询问：

$1,2,3,4,.....mex$ 是否都能由原数组构造出来，对于 $i$ 如果它可以由原数组构造出来，那么即有某个子数组 $mex(l,r)=i$，那么首先根据 $mex$ 的定义，我们可以知道 $[l,r]$ 上一定不能含有 $i$，其次，在这个基础上，$[l,r]$ 尽量大，才也有可能保证 $mex=i$，具体的你需要知道 $mex=i$ 的子数组的基本性质：

1. $mex=i$，则该序列不包括 $i$，否则可以拓展为 $mex=i+1$。

2. $mex$ 关于 $len$ 单调不降，即元素个数越多，$mex$ 越可能变大。

3. $mex$ 不包含 $i$，则 $mex\le i$，这个很显而易见。

其实这三条性质告诉了我们一种方式：

如果构造出 $mex=i$ 的可能方式？我们找到所以 $a_j=i$，以这些 $j$ 划分，它会形成若干个极长段。

为啥要选择两个 $i$ 之间的完整一段，这是基于第二点和第三点，显然不包括 $i$ 且最长的段才有可能构造出 $mex=i$。我们注意到枚举 $mex$ 这样的分割点总数至多为 $n$ 个，因为数组至多只有 $n$ 个位置，每个位置至多出现在某一种 $mex$ 中。也就是说至多有 $O(n)$ 段需要求 $mex$，这个就转化为了非常简单与经典的 $\mathrm{区}\mathrm{间}mex$ 询问。

解法一

普通莫队+值域分块平衡复杂度

建议先学习：值域分块系统级教学

对于莫队怎么求区间 $mex$，这个问题很简单。考虑权值域，我们每次莫队会往权值域内增加或者删除一个数，$cnt[x]++,cnt[x]--$，而最暴力地询问 $mex$ 则为从 $1$ 开始，一直到 $mex$，其中 $mex$ 是第一个 $cnt[mex]==0$ 的数。

那么这样的复杂度是，插入和删除是 $O(1)$，查询是 $O(n)$，如果换成树状数组的话，可以做到 $O(\log V)$ 插入和删除，$O(\log V)$ 查询。查询的话这里要说下如何使用线段树或者树状数组做到单 $\log V$ 的复杂度查询，我们只需要维护一个区间最小值，然后二分答案，找第一个 $cnt=0$ 的点，即为找最小值 $=0$ 的第一个点即可。

所以上述复杂度并不合理，我们需要 $O(1)$ 修改，查询可以要求劣一点，$O(\sqrt{V})$ 的查询就行了，这显然值域分块就能办到了，每次单点加值域块和值域点，查询枚举值域块找答案就行了。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #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 = 2e5 + 10;
int pos[N];
 
struct Mo
{
    int l, r;
 
    bool operator<(const Mo& other) const
    {
        return pos[l] ^ pos[other.l] ? pos[l] < pos[other.l] : pos[l] & 1 ? r < other.r : r > other.r;
    }
} node[N];
 
int n, q;
vector<int> t[N];
int a[N];
bool vis[N];
int val[N], valPos[N];
int siz, cnt, s[N], e[N];
int Cnt[N];
 
inline void update(const int x, const int v)
{
    const int id = pos[x];
    val[id] += v, valPos[x] += v;
}
 
inline void add(const int x)
{
    if (Cnt[x]++ == 0)update(x, 1);
}
 
inline void del(const int x)
{
    if (--Cnt[x] == 0)update(x, -1);
}
 
inline int queryMex()
{
    int id = 1;
    while (id <= cnt and val[id] == e[id] - s[id] + 1)id++;
    if (id > cnt)return n + 1;
    int ans = s[id];
    while (ans <= e[id] and valPos[ans])ans++;
    return ans;
}
 
inline void solve()
{
    cin >> n;
    siz = sqrt(n);
    cnt = (n + siz - 1) / siz;
    forn(i, 1, n)cin >> a[i], t[a[i]].push_back(i), pos[i] = (i - 1) / siz + 1;
    forn(i, 1, cnt)s[i] = (i - 1) * siz + 1, e[i] = i * siz;
    e[cnt] = n;
    forn(i, 1, n+1)
    {
        int l = 1;
        for (const auto r : t[i])
        {
            if (l == r)
            {
                l = r + 1;
                continue;
            }
            node[++q] = Mo(l, r - 1), l = r + 1;
        }
        if (l < n)node[++q] = Mo(l, n);
    }
    sortArr(node, q);
    int l = 1, r = 0;
    forn(i, 1, q)
    {
        const auto [L,R] = node[i];
        while (l < L)del(a[l++]);
        while (l > L)add(a[--l]);
        while (r < R)add(a[++r]);
        while (r > R)del(a[r--]);
        vis[queryMex()] = true;
    }
    int mex = 1;
    while (vis[mex])mex++;
    cout << mex;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    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;
}

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O(n\sqrt{n})$$

解法二

回滚莫队

太经典了，区间 $mex$ 问题回滚莫队也能解决，也是主要解决途径，具体的，我们观察到，如果往一个权值域增加一个数，那么我们的 $mex$ 显然需要暴力地更新到最新 $mex$，因为可能一开始已经有若干值了。

如图所示，如果 $+$ 上一个数，那么很容易发现，这个 $mex$ 是跳跃式更新，但是，删除如果一个数 消失了，且这个数 $<mex$，则 $mex=\mathrm{这}\mathrm{个}\mathrm{数}$，这个更新则是 $O(1)$ 的，所以我们可以使用只删不加的回滚莫队即可。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #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 = 2e5 + 10;
int pos[N];
 
struct Mo
{
    int l, r;
 
    bool operator<(const Mo& other) const
    {
        return pos[l] ^ pos[other.l] ? pos[l] < pos[other.l] : r > other.r;
    }
} node[N];
 
int n, q;
vector<int> t[N];
int a[N];
bool vis[N];
int curr = 1;
int siz, cnt, s[N], e[N];
int Cnt[N];
stack<int> back;
 
inline void del(const int x, const bool isBack = false)
{
    if (--Cnt[x] == 0 and x < curr)curr = x;
    if (isBack)back.push(x);
}
 
int tmp[N];
 
inline void solve()
{
    cin >> n;
    siz = sqrt(n);
    cnt = (n + siz - 1) / siz;
    forn(i, 1, n)cin >> a[i], t[a[i]].push_back(i), pos[i] = (i - 1) / siz + 1;
    forn(i, 1, cnt)s[i] = (i - 1) * siz + 1, e[i] = i * siz;
    e[cnt] = n;
    forn(i, 1, n+1)
    {
        int l = 1;
        for (const auto r : t[i])
        {
            if (l == r)
            {
                l = r + 1;
                continue;
            }
            node[++q] = Mo(l, r - 1), l = r + 1;
        }
        if (l < n)node[++q] = Mo(l, n);
    }
    sortArr(node, q);
    int l = 1, r = 0, last = 0;
    forn(i, 1, q)
    {
        const auto [L,R] = node[i];
        if (pos[L] == pos[R])
        {
            forn(i, L, R)tmp[a[i]]++;
            int mex = 1;
            while (tmp[mex])mex++;
            vis[mex] = true;
            forn(i, L, R)tmp[a[i]]--;
            continue;
        }
        if (pos[L] != last)
        {
            last = pos[L];
            curr = 1;
            forn(i, 1, n)Cnt[i] = 0;
            forn(i, s[pos[L]], n)Cnt[a[i]]++;
            while (Cnt[curr])curr++;
            l = s[pos[L]], r = n;
        }
        while (r > R)del(a[r--]);
        const int pre = curr;
        int tmpL = l;
        while (tmpL < L)del(a[tmpL++], true);
        vis[curr] = true;
        while (!back.empty())Cnt[back.top()]++, back.pop();
        curr = pre;
    }
    int mex = 1;
    while (vis[mex])mex++;
    cout << mex;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    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;
}

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O(n\sqrt{n})$$

解法三

在线主席树上二分

关于 $mex$ 和颜色数问题是属于同一条道路的，有一类经典分析，就是 $pre/last$ 分析。

对于询问区间 $[l,r]$ 的 $mex$，我们有这样的暴力方法：

枚举 $1\sim n$，然后 $check$ 它是否在这个区间内。这个 $check$ 它是否在一个区间内显然有很多种方法，我要讲一种很经典的方法，以便于拓展。对于一个数是否在这个区间内只需要看 $l\le last[x]\le r$，即我们从 $1\to r$ 时，$x$ 最近出现的位置需要在 $[l,r]$ 上，那么这个 $x$ 就对 $mex$ 有贡献。而 $mex$ 显然就为从 $1\sim n$ 第一个不满足 $last[x]\ge l$ 的点。

这个咋解决？我们抽象下：

为每个权值记录 $last$，对于每个查询 $[l,r]$：

我们维护 $\mathrm{权}\mathrm{值}\Rightarrow last[\mathrm{权}\mathrm{值}]$，而 $mex$ 即为 $1\sim n$ 上第一个 $last[i]<l$ 的 $i$，这个咋做？这不就是经典线段树二分吗？我们咋维护一个区间 $min$，我们就能找到 $last[i]<l$ 是哪一半边，就能二分出这个第一个点。完毕。

对于一堆查询，它们的 $r$ 并不固定，但每次查询的即为 $1\to r$ 更新完毕的状态，因为我们要保证 $last[x]$ 是更新到 $r$ 的最近的 $x$ 的位置。这玩意在线做，就是主席树了，查询 $root[r]$ 处的主席树，就是 $0\sim r$ 的状态了。当然也可以树套树直接暴力做，不过复杂度就劣了，我们要求查询的是前缀状态，主席树足矣。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #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 = 2e5 + 10;
int n, q;
vector<int> t[N];
int a[N];
bool vis[N];
int cnt;
constexpr int INF = 1e9 + 7;
 
struct Node
{
    int mi;
    int left, right;
} node[N << 5];
 
#define left(x) node[x].left
#define right(x) node[x].right
#define mi(x) node[x].mi
int root[N];
 
inline void pushUp(const int curr)
{
    mi(curr) = min(mi(left(curr)),mi(right(curr)));
}
 
inline void add(const int pre, int& curr, const int pos, const int val, const int l = 1, const int r = n + 2)
{
    node[curr = ++cnt] = node[pre];
    const int mid = l + r >> 1;
    if (l == r)
    {
        mi(curr) = val;
        return;
    }
    if (pos <= mid)add(left(pre),left(curr), pos, val, l, mid);
    else add(right(pre),right(curr), pos, val, mid + 1, r);
    pushUp(curr);
}
 
inline int queryMex(const int curr, const int pos, const int l = 1, const int r = n + 2)
{
    if (l == r)return l;
    const int mid = l + r >> 1;
    if (mi(left(curr)) < pos)return queryMex(left(curr), pos, l, mid);
    return queryMex(right(curr), pos, mid + 1, r);
}
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i], t[a[i]].push_back(i);
    forn(i, 1, n)add(root[i - 1], root[i], a[i], i);
    forn(i, 1, n+1)
    {
        int l = 1;
        for (const auto r : t[i])
        {
            if (l == r)
            {
                l = r + 1;
                continue;
            }
            vis[queryMex(root[r - 1], l)] = true, l = r + 1;
        }
        if (l < n)vis[queryMex(root[n], l)] = true;
    }
    int mex = 1;
    while (vis[mex])mex++;
    cout << mex;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    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;
}

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O(n\log n)$$

第四种解法

离线扫描线询问+线段树上二分

既然本题可以在线做，当然也可以离线做，我们把所有询问挂载在序列扫描线上，直接从 $1-r$ 更新线段树的同时处理询问即可。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #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 = 2e5 + 10;
int n, q;
vector<int> t[N];
int a[N];
bool vis[N];
int cnt;
constexpr int INF = 1e9 + 7;
 
struct Node
{
    int mi;
} node[N << 2];
 
#define mi(x) node[x].mi
 
inline void push_up(const int curr)
{
    mi(curr) = min(mi(ls(curr)),mi(rs(curr)));
}
 
inline void update(const int curr, const int pos, const int val, const int l = 1, const int r = n + 2)
{
    if (l == r)
    {
        mi(curr) = val;
        return;
    }
    const int mid = l + r >> 1;
    if (pos <= mid)update(ls(curr), pos, val, l, mid);
    else update(rs(curr), pos, val, mid + 1, r);
    push_up(curr);
}
 
inline int queryMex(const int curr, const int pos, const int l = 1, const int r = n + 2)
{
    if (l == r)return l;
    const int mid = l + r >> 1;
    if (mi(ls(curr)) < pos)return queryMex(ls(curr), pos, l, mid);
    return queryMex(rs(curr), pos, mid + 1, r);
}
 
vector<int> seg[N];
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i], t[a[i]].push_back(i);
    forn(i, 1, n+1)
    {
        int l = 1;
        for (const auto r : t[i])
        {
            if (l == r)
            {
                l = r + 1;
                continue;
            }
            seg[r - 1].push_back(l), l = r + 1;
        }
        if (l < n)seg[n].push_back(l);
    }
    forn(r, 1, n)
    {
        update(1, a[r], r);
        for (const auto l : seg[r])vis[queryMex(1, l)] = true;
    }
    int mex = 1;
    while (vis[mex])mex++;
    cout << mex;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    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;
}

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O(n\log n)$$

一些细节

本题注意 $mex$ 最大为 $n+1$，所以记得枚举到 $n+1$。注意每个包含 $i$ 的位置是不可取的，我们可以适当的剪枝，对于 $[l,r]$ 我们要求这个区间的 $mex$ 为 $i$，如果区间元素数量都不够，那显然不可能形成 $i$。最后，当然不用完全枚举 $1\sim n+1$，一直枚举到 $mex$ 确定的时候就可以停止枚举了。

剪枝参照

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #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 = 2e5 + 10;
int n, q;
vector<int> t[N];
int a[N];
bool vis[N];
int cnt;
constexpr int INF = 1e9 + 7;
 
struct Node
{
    int mi;
} node[N << 2];
 
#define mi(x) node[x].mi
 
inline void push_up(const int curr)
{
    mi(curr) = min(mi(ls(curr)),mi(rs(curr)));
}
 
inline void update(const int curr, const int pos, const int val, const int l = 1, const int r = n + 2)
{
    if (l == r)
    {
        mi(curr) = val;
        return;
    }
    const int mid = l + r >> 1;
    if (pos <= mid)update(ls(curr), pos, val, l, mid);
    else update(rs(curr), pos, val, mid + 1, r);
    push_up(curr);
}
 
inline int queryMex(const int curr, const int pos, const int l = 1, const int r = n + 2)
{
    if (l == r)return l;
    const int mid = l + r >> 1;
    if (mi(ls(curr)) < pos)return queryMex(ls(curr), pos, l, mid);
    return queryMex(rs(curr), pos, mid + 1, r);
}
 
vector<int> seg[N];
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i], t[a[i]].push_back(i);
    forn(i, 1, n+1)
    {
        int l = 1;
        bool isAdd = false;
        for (const auto r : t[i])
        {
            if (l == r)
            {
                l = r + 1;
                continue;
            }
            if (r - l >= i - 1)
            {
                isAdd = true;
                seg[r - 1].push_back(l);
            }
            l = r + 1;
        }
        if (l < n)
        {
            if (n - l + 1 >= i - 1)
            {
                isAdd = true;
                seg[n].push_back(l);
            }
        }
        if (!isAdd)break;
    }
    forn(r, 1, n)
    {
        update(1, a[r], r);
        for (const auto l : seg[r])vis[queryMex(1, l)] = true;
    }
    int mex = 1;
    while (vis[mex])mex++;
    cout << mex;
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    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;
}