传送门

A Number Transformation

赛时做复杂了，其实就是两种情况，x大于y或者x不能整除y的情况下不行，能整出的情况下直接输出1和x/y即可写复杂了，得亏数据范围小不然感觉容易挂

//
//  Author : north_h
//  File : A.cpp
//  Time : 2023/7/18/12:56
//

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;

using namespace std;


void solve() {
    int x, y;
    cin >> x >> y;
    if (y % x != 0 || x > y)cout << "0 0" << endl;
    else if (x == y)cout << "1 1" << endl;
    else {
        int xx = y / x;
        for (int i = 2; i <= xx; i++) {
            int sum = 1;
            int b = 0;
            while (sum < xx)sum *= i, b++;
            if (sum == xx) {
                cout << b << ' ' << i << endl;
                return;
            }
        }
    }
}

int main() {
    IOS;
    int h_h = 1;
    cin >> h_h;
    while (h_h--)solve();
    return 0;
}

B Dictionary

直接根据给出的形式计算即可，可以直接不思考预处理出来直接查询

//
//  Author : north_h
//  File : B.cpp
//  Time : 2023/7/18/13:12
//

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;

using namespace std;


void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        int cnt = (int) (s[0] - 'a') * 25 + s[1] - 'a' + 1;
        if (s[1] > s[0])cnt--;
        cout << cnt << endl;
    }
}

int main() {
    IOS;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--)solve();
    return 0;
}

C Infinite Replacement

这个题真的时排列组合的数量都忘了，还自己一个一个去数，直接就是2 的次方，一个秒了的

//
//  Author : north_h
//  File : C.cpp
//  Time : 2023/7/18/13:32
//

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;

using namespace std;

void solve() {
    string a, b;
    cin >> a >> b;
    bool ok = false;
    for (auto i: b)if (i == 'a')ok = true;
    if (ok && b.size() == 1)cout << 1 << endl;
    else if (ok && b.size())cout << -1 << endl;
    else cout << (1ll << a.size()) << endl;
}

int main() {
    IOS;
    int h_h = 1;
    cin >> h_h;
    while (h_h--)solve();
    return 0;
}

D A-B-C Sort

可以发现两种操作以后只是改变一对一对相邻的位置关系，奇数时第一个不参与偶数时从第一个开始，后面的一对一对的排序，最后判断是否是非递减的即可，直接写挂了，难受

//
//  Author : north_h
//  File : D.cpp
//  Time : 2023/7/18/14:45
//

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se seco1nd
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;

using namespace std;


void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for(auto &i:a)cin >>i;
    int last = 0;
    if (n % 2 != 0)last = a[0];
    //cout<<last<<endl;
    for (int i = (n % 2 != 0); i < n; i += 2) {
        if (min(a[i], a[i + 1]) >= last)last = max(a[i], a[i + 1]);
        else {
            cout << "NO" << endl;
            return;
        }
    }
    cout << "YES" << endl;
}

int main() {
    IOS;
    int h_h = 1;
    cin >> h_h;
    while (h_h--)solve();
    return 0;
}

F Desktop Rearrangement

主要还是读懂题意，读懂题意以后就是动态的前缀和，单点修改区间查询，所以可以使用树状数组来做，可以用二维的树状数组来做，因为这道题的原因，也可以吧坐标映射到一维上，用一维的树状数组来做

//
//  Author : north_h
//  File : E.cpp
//  Time : 2023/7/18/19:06
//                  _   _         _     
// _ __   ___  _ __| |_| |__     | |__  
//| '_ \ / _ \| '__| __| '_ \    | '_ \ 
//| | | | (_) | |  | |_| | | |   | | | |
//|_| |_|\___/|_|   \__|_| |_|___|_| |_|
//                          |_____|     

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define endl '\n'
const int N = 1010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;

using namespace std;

int tr[N][N];
char g[N][N];
int n, m, q;

inline int lowbit(int x) {
    return x & (-x);
}

void add(int x,int y,int k) {
    for (int i = x; i <= n; i += lowbit(i)) {
        for (int j = y; j <= m; j += lowbit(j)) {
            tr[i][j] += k;
        }
    }
}

int query(int x,int y) {
    int res = 0;
    for (int i = x; i; i -= lowbit(i)) {
        for (int j = y; j; j -= lowbit(j)) {
            res += tr[i][j];
        }
    }
    return res;
}

int query(int x1,int y1,int x2,int y2){
    return query(x2,y2)-query(x2,y1-1)-query(x1-1,y2)+query(x1-1,y1-1);
}

void solve() {
    cin >> n >> m >> q;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> g[i][j];
            if (g[i][j] == '*') {
                sum++;
                add(i, j, 1);
            }
        }
    }
    while (q--) {
        int x, y;
        cin >> x >> y;
        if (g[x][y] == '*') {
            sum--;
            g[x][y] = '.';
            add(x, y, -1);
        } else {
            sum++;
            g[x][y] = '*';
            add(x, y, 1);
        }
        int col = sum / n;
        int cnt = sum % n;
        int t = 0;
        if (col)t += query(1, 1, n, col);
        if (cnt)t += query(1, col + 1, cnt, col + 1);
        cout << sum - t << endl;
    }
}

int main() {
    IOS;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--)solve();
    return 0;
}