黃綠題選刷

Star_F發表於2024-10-23

  • [ABC376D] Cycle

找到包含節點 1 的環，直接從節點一出發，BFS，如果第二次遍歷到了節點1，直接輸出時間即可。

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
inline int rd(){
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
#define rd rd()

void wt(int x){
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        wt(x / 10);
    putchar(x % 10 + '0');
    return;
}
void wt(int x, char k){
    wt(x),putchar(k);
}

namespace Star_F{
    const int N = 200005;
    int n, m;
    bool vis[N], f;
    
    int h[N],e[N],ne[N],idx;
    void add(int a,int b){
        e[++idx] = b, ne[idx] = h[a], h[a] = idx;
    }

    void bfs(){
        queue<PII> q;
        q.push({1, 0});
        while(!q.empty()){
            int u  = q.front().fi,ti=q.front().se;

            q.pop();
            for (int i = h[u]; i;i=ne[i]){
                int j = e[i];
                if(j==1){
                    cout << ti + 1 << endl;
                    f = 1;
                    return;
                }
                if(!vis[j]){
                    q.push({j,ti + 1});
                    vis[j] = 1;
                }
            }
        }
    } 
    void Main(){
        n = rd, m = rd;
        FOR(i,1,m){
            int a, b;
            a = rd, b = rd;
            add(a, b);
        }

        
        bfs();
        if(!f) cout << -1;
    }

}

signed main(){
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ClockA;
    int T=1;
    // T=rd;
    while(T--) Star_F::Main();
    // ClockB;
    return 0;
}


/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
* \- ▌           Name: Star_F              ▀ ▀
*  - ▌                            (o)          ▀
* /- ▌            Go Go Go !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

  • [ABC348F] Oddly Similar

其實就是個暴力，用 bitset 最佳化一下就行。
先寫出用 bool 陣列能過的程式碼，再把 bool陣列 最後一維用 bitset 最佳化掉。

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
inline int rd(){
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
#define rd rd()

void wt(int x){
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        wt(x / 10);
    putchar(x % 10 + '0');
    return;
}
void wt(int x, char k){
    wt(x),putchar(k);
}

namespace Star_F{
    const int N = 2001, M = 1000;
    int n, m, a[N][N], res;
    bitset<N> cnt[N][M], f;
    void Main(){
        n = rd, m = rd;
        for (int i = 1; i <= n; ++ i )
            for (int j = 1; j <= m; ++ j ){
                a[i][j] = rd;
                cnt[j][a[i][j]][i] = 1;
            }
        for (int i = 1; i <= n; ++ i ) {
            f.reset();
            for (int j = 1; j <= m; ++ j ) f ^= cnt[j][a[i][j]];
            f[i] = 0;		 
            res += f.count();
        }
        wt(res/2, '\n');
    }
}

signed main(){
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ClockA;
    int T=1;
    // T=rd;
    while(T--) Star_F::Main();
    // ClockB;
    return 0;
}


/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
* \- ▌           Name: Star_F              ▀ ▀
*  - ▌                            (o)          ▀
* /- ▌            Go Go Go !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

  • [ABC349D] Divide Interval

挺簡單的一道題，貪心。
每次肯定是能跑多遠跑多遠，把 \(2^i\) 打表出來，從大到小列舉，再判斷能否整除。

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;

#define int long long
inline int rd(){
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
#define rd rd()

void wt(int x){
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        wt(x / 10);
    putchar(x % 10 + '0');
    return;
}
void wt(char x){
    putchar(x);
}
void wt(int x, char k){
    wt(x),putchar(k);
}

namespace Star_F{
    const int N = 10000005;
    ll l,r,h1[N],h2[N],ans,a[N];
    void Main(){
        l = rd, r = rd;
        a[0] = 1;
        FOR(i, 1, 62) a[i] = a[i - 1] << 1ll;
        while (l != r){
            ROF(i,60,0){
                if (l + a[i] <= r && l % a[i] == 0){
                    h1[++ans] = l;
                    h2[ans] = l + a[i];
                    l += a[i];
                    break;
                }
            }
        }
        wt(ans, '\n');
        FOR(i,1,ans) cout<<h1[i]<<" "<<h2[i]<<endl;
    }

}

signed main(){
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ClockA;
    int T=1;
    // T=rd;
    while(T--) Star_F::Main();
    // ClockB;
    return 0;
}


/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
* \- ▌           Name: Star_F              ▀ ▀
*  - ▌                            (o)          ▀
* /- ▌            Go Go Go !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

  • [AGC032B] Balanced Neighbors

考慮一個圖滿足

  • 不連通
  • 每個點和這個點的臨點的和為 \(S\)

那麼這個圖的補圖滿足

  • 聯通
  • 每個點的所有的鄰點的和為 \(T\)（因為頂點總和是不變的，所以 \(T=總頂點和 - S\),而S \(S\) 也是不變的。

image

構造不連通，且每個點和這個點的臨點的和為 \(S\) 的圖很容易 構造，按照 \(n\) 的奇偶分類即可。

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair<int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;

// 輸入函式
inline int rd() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

void wt(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) wt(x / 10);
    putchar(x % 10 + '0');
}

void wt(char x) {
    putchar(x);
}

void wt(int x, char k) {
    wt(x), putchar(k);
}

namespace Star_F {
    void Main() {
        int n = rd(); 

      
        printf("%d\n", n * (n - 1) / 2 - n / 2);

      
        FOR(i, 1, n) {
            FOR(j, i + 1, n) {
                if (n & 1) { 
                    if (i + j != n) { 
                        printf("%d %d\n", i, j); 
                    }
                } else { 
                    if (i + j != n + 1) { 
                        printf("%d %d\n", i, j); 
                    }
                }
            }
        }
    }
}

signed main() {
    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);
    ClockA;
    int T = 1;
    // T = rd();
    while (T--) Star_F::Main();
    ClockB;
    return 0;
}

/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
* \- ▌           Name: Star_F              ▀ ▀
*  - ▌                            (o)          ▀
* /- ▌            Go Go Go !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

  • P11208 『STA - R8』輪迴瘋狂

只用操作二，最多進行 \(n-1\) 次即可完成，所一答案最大為 \(n-1\)

考慮操作一怎麼使用：
可以考慮反著想，操作二是每次從大到小增加一個數，增加的數肯定是當前序列中的最小值（因為是從大到小新增），然後對這個選的數的位置進行插入排序，看看是否會超過 \(n-1\) 。

具體看程式碼吧

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int n, a[N], p[N], b[N], ans;

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        p[a[i]] = i;
    }

    ans = n;
    int S = 0, m = 0;
    for (int i = n; i >= 1; --i) {
        b[++m] = p[i];
        for (int j = m; j >= 2; --j) {
            if (b[j] < b[j - 1]) break;
            swap(b[j], b[j - 1]);
            ++S;
        }
        if (S > n) break;
        ans = min(ans, S + i - 1);
    }

    cout << ans << '\n';
    return 0;
}

  • P11206 「Cfz Round 9」Dove

輸入一棵樹，且只有 n−1 條邊，所以，肯定無向無環，可以用 DFS 來搜尋。
由於限制，所以值為 1 和 n 的點之間應該有一條邊。且值為 n 的點只能與 1 相連。
以此類推，儘量將大值放到度為 1的點上.將小值放到與它相鄰的點上。

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;

inline int rd(){
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
#define rd rd()

void wt(int x){
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        wt(x / 10);
    putchar(x % 10 + '0');
    return;
}
void wt(char x){
    putchar(x);
}
void wt(int x, char k){
    wt(x),putchar(k);
}

namespace Star_F {
    const int N = 1e5 + 5;
    vector<int> mp[N];
    int n, l, r, num[N];

    void dfs(int u, int f) {
        for (int v : mp[u]) {
            if (v == f) continue;
            dfs(v, u);
        }
        if (!num[u]) {
            num[u] = r--;
            if (f && !num[f]) num[f] = l++;
        }
    }

    void Main() {
        n = rd;
        for (int i = 1; i <= n; i++) mp[i].clear(), num[i] = 0;
        for (int i = 2; i <= n; i++) {
            int u = rd, v = rd;
            mp[u].emplace_back(v);
            mp[v].emplace_back(u);
        }
        l = 1, r = n;
        dfs(1, 0);
        for (int i = 1; i <= n; i++)
            wt(num[i], ' ');
        wt('\n');
    }
}

signed main() {
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ClockA;
    int T=1;
    T=rd;
    while(T--) Star_F::Main();
    // ClockB;
    return 0;
}

/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
* \- ▌           Name: Star_F              ▀ ▀
*  - ▌                            (o)          ▀
* /- ▌            Go Go Go !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

  • P11129 【MX-X5-T1】「GFOI Round 1」Inverted World

打表題，可以爆搜找通項公式
當然也可以：

題解 ----- cff_0102

  • P11205 「Cfz Round 9」Hope

挺簡單的，先排個序，然後肯定儘可能的讓序列變成 \(1,2,3,...n-1,n\) ，如果某個 \(a_i < i\) ，那就不管他，加入備選集合。否則加入必選集合
如果 \(必選集合總和 <= 備選集合元素數量+1 && 必選集合總和 + 備選集合總和 >= 備選集合元素數量+1\) 那麼答案加一（通俗的說就是透過刪除的數能湊出來一個更大的）

點選檢視程式碼 
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
#define ll long long
typedef pair<int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
#define int long long
inline int rd() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

#define rd rd()

void wt(int x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        wt(x / 10);
    putchar(x % 10 + '0');
    return;
}
void wt(char x) {
    putchar(x);
}
void wt(int x, char k) {
    wt(x), putchar(k);
}

namespace Star_F {
    const int N = 100005;
    int a[N];
    void Main() {
        int n = rd, A = 0, B = 0, f = 1;
        FOR(i,1,n) a[i] = rd; 
        sort(a + 1, a + 1 + n);
        FOR(i,1,n)
            if (a[i] >= f) A += a[i] - (f++);
            else B += a[i] - 1;
        wt(f+(A <= f && A + B >= f),'\n');
    }
}

signed main() {
    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);
    ClockA;
    int T = 1;
    T = rd;
    while (T--) Star_F::Main();
    // ClockB;
    return 0;
}

/*
*          ▀▀▀██████▄▄▄       _______________
*          ▄▄▄▄▄  █████████▄  /                 \
*         ▀▀▀▀█████▌ ▀▐▄ ▀▐█ | Code has no BUG!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀
*/

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