【刷题】黄绿题选刷

• [ABC376D] Cycle

找到包含节点 1 的环，直接从节点一出发，BFS，如果第二次遍历到了节点1，直接输出时间即可。

点击查看代码

#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

其实就是个暴力，用 bitset 优化一下就行。
先写出用 bool 数组能过的代码，再把 bool数组 最后一维用 bitset 优化掉。

点击查看代码

#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

挺简单的一道题，贪心。
每次肯定是能跑多远跑多远，把 \(2^i\) 打表出来，从大到小枚举，再判断能否整除。

点击查看代码

#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

考虑一个图满足

• 不连通
• 每个点和这个点的临点的和为 \(S\)

那么这个图的补图满足

• 联通
• 每个点的所有的邻点的和为 \(T\)（因为顶点总和是不变的，所以 \(T=总顶点和 - S\),而S \(S\) 也是不变的。

构造不连通，且每个点和这个点的临点的和为 \(S\) 的图很容易 构造，按照 \(n\) 的奇偶分类即可。

点击查看代码

#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;

// 输入函数
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』轮回疯狂

只用操作二，最多进行 \(n-1\) 次即可完成，所一答案最大为 \(n-1\)

考虑操作一怎么使用：
可以考虑反着想，操作二是每次从大到小增加一个数，增加的数肯定是当前序列中的最小值（因为是从大到小添加），然后对这个选的数的位置进行插入排序，看看是否会超过 \(n-1\) 。

具体看代码吧

点击查看代码

#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

输入一棵树，且只有 n−1 条边，所以，肯定无向无环，可以用 DFS 来搜索。
由于限制，所以值为 1 和 n 的点之间应该有一条边。且值为 n 的点只能与 1 相连。
以此类推，尽量将大值放到度为 1的点上.将小值放到与它相邻的点上。

点击查看代码

#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

打表题，可以爆搜找通项公式
当然也可以：

题解 ----- cff_0102

• P11205 「Cfz Round 9」Hope

挺简单的，先排个序，然后肯定尽可能的让序列变成 \(1,2,3,...n-1,n\) ，如果某个 \(a_i < i\) ，那就不管他，加入备选集合。否则加入必选集合
如果 \(必选集合总和 <= 备选集合元素数量+1 && 必选集合总和 + 备选集合总和 >= 备选集合元素数量+1\) 那么答案加一（通俗的说就是通过删除的数能凑出来一个更大的）

点击查看代码

#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!  |
*      ▀▀█████▄▄ ▀██████▄██ | _________________/
*       ▀▄▄▄▄▄  ▀▀█▄▀█════█▀ |/
*            ▀▀▀▄  ▀▀███ ▀       ▄▄
*         ▄███▀▀██▄████████▄ ▄▀▀▀▀▀▀█▌   ______________________________
*       ██▀▄▄▄██▀▄███▀ ▀▀████      ▄██  █                               \\
*    ▄▀▀▀▄██▄▀▀▌████▒▒▒▒▒▒███     ▌▄▄▀▀▀▀█_____________________________ //
*    ▌    ▐▀████▐███▒▒▒▒▒▐██▌
*    ▀▄▄▄▄▀   ▀▀████▒▒▒▒▄██▀
*              ▀▀█████████▀
*            ▄▄██▀██████▀█
*          ▄██▀     ▀▀▀  █
*         ▄█             ▐▌
*     ▄▄▄▄█▌              ▀█▄▄▄▄▀▀▄
*    ▌     ▐                ▀▀▄▄▄▀
*     ▀▀▄▄▀     ██
* \  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀
*/

• P11148 [THUWC 2018] 零食

贪心的想，肯定要把小的数两两消除。
枚举消除的长度，两种零食消除的长度可能相差一，再分别计算即可。
（维护一个后缀和，直接 \(\mathcal{O}(1)\) 计算即可。

点击查看代码

#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
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 n, m, a[N], b[N],A[N], B[N];
    int f__a(int x){
        if(x<=n) return A[x];
        return 0;
    }
    int f__b(int x){
        if(x<=m) return B[x];
        return 0;
    }
    void Main(){
        n = rd;
        FOR(i,1,n) a[i] = rd;
        m = rd;
        FOR(i,1,m) b[i] = rd;
        sort(a + 1, a + n + 1);
        sort(b + 1, b + m + 1);

        A[n + 1] = B[m + 1] = 0;
        ROF(i,n,1)
            A[i] = A[i + 1] + a[i];
        ROF(i,m,1)
            B[i] = B[i + 1] + b[i];
        int ans = -0x7fffffffffffffff;
        FOR(i,1,min(n,m)){
            if (i <= n && i <= m)
                ans = max(f__a(i + 1) + f__b(i + 1), ans);
            if (i + 1 <= n && i <= m)
                ans = max(f__a(i + 2) + f__b(i + 1), ans);
            if (i <= n && i + 1 <= m)
                ans = max(f__a(i + 1) + f__b(i + 2), ans);
        }
        wt(ans, '\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 !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

• P11209 『STA - R8』小熊游景点 II

看到 异或 操作考虑字典树。
遍历 \(k\) 的每一位，在分讨 \(a,b\) 这位的取值，什么时候能产生贡献，分别计算即可。

点击查看代码

#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
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 = 1e6 + 5;
    int n, q, a[N], b[N];
    struct Tree{
        int tr[N * 30][2], cnt[N * 30][4], ed[N * 30], cc;

        void insert(int x,int y){
            int u = 0;
            for (int i = 30; i >= 0;i--){
                int c = x >> i & 1, d = y >> i & 1;
                cnt[u][c * 2 + d]++;
                if(!tr[u][c^d])
                    tr[u][c ^ d] = ++cc;
                u = tr[u][c ^ d];
            }
            ed[u]++;
        }

        int query(int x){
            int u = 0, ans = 0;
            bool f = 1;
            for (int i = 30; i >= 0;i--){
                int c = x >> i & 1;
                ans += ed[u] + cnt[u][c << 1 | 1];
                if(tr[u][c])
                    u = tr[u][c];
                else{
                    f = 0;
                    break;
                }
            }
            if(f)
                ans += ed[u];
            return ans;
        }
    } T;

    void Main(){

        int t;
        t = rd, n = rd, q = rd;
        FOR(i, 1, n) a[i] = rd;
        FOR(i, 1, n) b[i] = rd;
        FOR(i, 1, n) T.insert(a[i], b[i]);

        int k = 0, lst = 0;
        while(q--){
            k = rd;
            k ^= lst * t;
            wt(lst=T.query(k), '\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 !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

• P11157 【MX-X6-T3】さよならワンダーランド

分析公式题。
看到有大量重复且复杂的式子，考虑换元。
一大堆不等式条件，考虑转化为最关键元素的上下界。

本题就是缩小上下界，然后维护一个后缀最小值进行判断。

• P11130 【MX-X5-T2】「GFOI Round 1」Interstellar

感觉挺水的。

设 \(d = m/n\)，然后 \(n\) 每次乘以 \(gcd(n,d)\) ，相应的，\(n*=gcd(n,d),d/=gcd(n,d)\)

点击查看代码

#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
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 = 300005;
    
    
    void Main(){
        int n = rd, m = rd;
        if(m%n)
            wt(-1, '\n');
        else{
            int d = m / n, ans = 0;
            while(1){
                int g = __gcd(n, d);
                if(g==1)
                    break;
                d /= g, n *= g, ans++;
            }
            if(d>1)
                wt(-1, '\n');
            else
                wt(ans, '\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 !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

• P11215 【MX-J8-T3】水星湖

DFS，没啥好说的，难点在代码。

点击查看代码

#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
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 M = 3005,N=100005;
    const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
    int a[M][M], n, m, q, r, k, ans;
    int t[M][M]; 
    int sx[N], sy[N], snum; 

    bool f(int x, int y)  {
        FOR(i, 0, 3) {
            int nx = x + dx[i], ny = y + dy[i];
            if (a[nx][ny] == 1) return 1;
        }
        return 0;
    }

    void dfs(int x, int y) {
        FOR(i, 0, 3) {
            int nx = x + dx[i], ny = y + dy[i];
            if (nx >= 1 && nx <= n && ny >= 1 && ny <= m)
                if (f(nx, ny) && !a[nx][ny]) {
                    a[nx][ny] = 2;
                    t[nx][ny] = t[x][y] + 1;
                    dfs(nx, ny);
                }
        }
    }

    void Main() {
        n = rd, m = rd, q = rd, r = rd, k = rd;
        FOR(i, 1, q) {
            int lx, ly, rx, ry;
            lx = rd, ly = rd, rx = rd, ry = rd;
            FOR(col, lx, rx)
                FOR(row, ly, ry)
                    a[col][row] = 1; 
        }
        FOR(i, 1, r) {
            int ti, x, y;
            ti = rd, x = rd, y = rd;
            t[x][y] = ti; 
            a[x][y] = 2;
            if (!f(x, y))
                sx[++snum] = x, sy[snum] = y;  
            dfs(x, y);
        }
        FOR(j, 1, snum) {
            int x = sx[j], y = sy[j];
            bool flag = 0;
            FOR(i, 0, 3) {
                int nx = x + dx[i], ny = y + dy[i];
                if (a[nx][ny] == 2 && t[nx][ny] - t[x][y] <= k) {
                    flag = 1;
                    break;
                }
            }
            if (!flag) a[x][y] = 0;
        }
        FOR(i, 1, n)
            FOR(j, 1, m)
                if (a[i][j] == 2) ans++;
        wt(ans, '\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 !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

• P11207 「Cfz Round 9」Rose

首先观察到答案不超过2。
判断答案为0直接模拟一变过程即可。
难点在于如何判断答案为 1：

题解

点击查看代码

#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
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 long long N = 2e5 + 10;
    int q, n, a[N];
    string s1, s2;
    map<char, int> pa;

    void Main() {
        
        pa['P'] = 1;
        pa['W'] = 2;
        pa['V'] = 3;
        cin >> q;

        while (q--) {
            cin >> n >> s1 >> s2;
            n <<= 1;

            FOR(i, 1, n) {
                if (i % 2) a[i] = pa[s1[i >> 1]];
                else a[i] = pa[s2[(i >> 1) - 1]];
            }

            int cnt1 = 0, cnt2 = 0, cnt3 = 0, p = N;
            bool f = 0;

            FOR(i, 1, n) {
                if (a[i] == 1) cnt1++;
                else if (a[i] == 2) cnt2++;
                else cnt3++;

                if (cnt1 == cnt2 && cnt2 == cnt3) {
                    if (i % 2) {
                        wt(0, '\n');
                        f = 1;
                    }
                    p = i;
                    break;
                }
            }

            if (f) continue;

            cnt1 = cnt2 = cnt3 = 0;
            int t12 = 0, t13 = 0, t21 = 0, t23 = 0, t31 = 0, t32 = 0;

            for (int i = 1; i <= n; i += 2) {
                if (a[i] == 1) cnt1++;
                else if (a[i] == 2) cnt2++;
                else cnt3++;

                if (cnt1 - 1 == cnt2 && cnt2 - 1 == cnt3 && p > t12) {
                    wt(1, '\n');
                    f = 1;
                    break;
                } else if (cnt1 - 1 == cnt3 && cnt3 - 1 == cnt2 && p > t13) {
                    wt(1, '\n');
                    f = 1;
                    break;
                } else if (cnt2 - 1 == cnt1 && cnt1 - 1 == cnt3 && p > t21) {
                    wt(1, '\n');
                    f = 1;
                    break;
                } else if (cnt2 - 1 == cnt3 && cnt3 - 1 == cnt1 && p > t23) {
                    wt(1, '\n');
                    f = 1;
                    break;
                } else if (cnt3 - 1 == cnt1 && cnt1 - 1 == cnt2 && p > t31) {
                    wt(1, '\n');
                    f = 1;
                    break;
                } else if (cnt3 - 1 == cnt2 && cnt2 - 1 == cnt1 && p > t32) {
                    wt(1, '\n');
                    f = 1;
                    break;
                }

                if (a[i + 1] == 1) cnt1++;
                else if (a[i + 1] == 2) cnt2++;
                else cnt3++;

                if (cnt1 - 1 == cnt2 && cnt2 - 1 == cnt3) t12 = i;
                else if (cnt1 - 1 == cnt3 && cnt3 - 1 == cnt2) t13 = i;
                else if (cnt2 - 1 == cnt1 && cnt1 - 1 == cnt3) t21 = i;
                else if (cnt2 - 1 == cnt3 && cnt3 - 1 == cnt1) t23 = i;
                else if (cnt3 - 1 == cnt1 && cnt1 - 1 == cnt2) t31 = i;
                else if (cnt3 - 1 == cnt2 && cnt2 - 1 == cnt1) t32 = i;
            }

            if (!f) wt(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 !               ▀ ▀
* /  ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ ▀
*/

• P11036 【MX-X3-T3】「RiOI-4」GCD 与 LCM 问题

很奇怪的构造法

题解

点击查看代码

#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
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 {

    int a, b, c, d;

    void solve() {
        int cnt = 0, nc = a;
        while (nc % 2 == 0) nc /= 2, cnt++;
        c = pow(2, cnt + 1), d = a + c;
        cout << 1 << " " << c << " " << d << '\n';
    }

    void Main() {
        int T = rd;
        while (T--) {
            a = rd;
            solve();
        }
    }

}

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

• P11008 『STA - R7』异或生成序列

发现只要确定了 \(p1\)，其他的都可以地推出来。
这样做时间复杂度是 \(\mathcal{O}(n^2)\)
可以发现 \(p[i]\) 是大于 \(0\) 的，所以 ，\(sum[i]^p[1] != 0\)，也就是说 \(p[1]!=sum[i]\)，（$sum[i] 表示前 \(i\) 个 \(b[i]\) 异或起来。

所以可以进行剪枝，时间复杂度 \(\mathcal{O}(n)\)