A - 2UP3DOWN

#include<bits/stdc++.h>

using namespace std;

#define int long long

void solve() {
    int a, b;
    cin >> a >> b;
    if (a < b and b - a <= 2)
        cout << "Yes\n";
    else if (a > b and a - b <= 3)
        cout << "Yes\n";
    else
        cout << "No\n";

    return;
}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    solve();
    return 0;
}

B - 326-like Numbers

#include<bits/stdc++.h>

using namespace std;

#define int long long

void solve() {

}

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    for( int a , b , c; n ; n ++ ){
        a = n % 10 , b = ( n / 10 ) % 10 , c =  n / 100;
        if( b * c == a ){
            cout << n << "\n";
            return 0;
        }
    }
    return 0;
}

C - Peak

#include<bits/stdc++.h>

using namespace std;

#define int long long


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (auto &i: a) cin >> i;
    sort(a.begin(), a.end());
    int res = 0;
    for( int l = 0 , r = 0 ; l < n ; l ++ ){
        while( r+1 < n and a[r+1] - a[l] < m ) r ++;
        res = max( res , r - l + 1);
    }
    cout << res << "\n";
    return 0;
}

D - ABC Puzzle

因为n 的范围很小，所以可以直接暴搜过去，加一些剪枝就能跑的飞快

#include <bits/stdc++.h>

using namespace std;

#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e10, INF = 1e18;
const int mod = 998244353;

const int N = 1e6;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    string r, c;
    cin >> n >> r >> c;
    vector g(n, vector<char>(n, '.'));
    vi visR(n), visC(n);

    auto dfs = [&](auto &&self, int x, int y) -> void {
        if (y == n) {
            if (visR[x] < 7) return;
            self(self, x + 1, 0);
            return;
        }
        if (x == n) {
            bool flag = true;
            for (int i = 0; flag and i < n; i++)
                if (visR[i] != 7 or visC[i] != 7) flag = false;
            for (int i = 0; flag and i < n; i++) {
                for (int j = 0; flag and j < n; j++) {
                    if (g[i][j] == '.') continue;
                    if (g[i][j] != r[i]) flag = false;
                    break;
                }
                for (int j = 0; flag and j < n; j++) {
                    if (g[j][i] == '.') continue;
                    if (g[j][i] != c[i]) flag = false;
                    break;
                }
            }

            if (flag) {
                cout << "Yes\n";
                for (auto it: g) {
                    for (auto c: it) cout << c;
                    cout << "\n";
                }
                exit(0);
            }
            return;
        }

        self(self, x, y + 1);
        for (int i = 0; i < 3; i++) {
            if ((visR[x] & (1 << i)) or (visC[y] & (1 << i))) continue;
            visR[x] ^= (1 << i), visC[y] ^= (1 << i);
            g[x][y] = char(i + 'A');
            self(self, x, y + 1);
            visR[x] ^= (1 << i), visC[y] ^= (1 << i);
            g[x][y] = '.';
        }

    };

    dfs(dfs, 0, 0);
    cout << "No\n";
    return 0;
}

E - Revenge of "The Salary of AtCoder Inc."

当前状态是$x$，$f[x]$表示当前掷出值$x$为，并继到游戏结束期望获得的钱数。

显然$f[n]=0$

对于当前状态$x$，有两种情况

$\frac x n$掷出$y\le x $，游戏结束，后继收入为$0$
对于$i>x$，有$\frac 1 n $掷出$i$，收入为$a_i$，后继收入为$f[i]$

所以$f[x]=\frac x n \times 0 + \sum_{i=x+1}^{n} \frac 1 n ( f[i] + a[i] )$

所以倒序枚举状态，在后缀和统计一下可以$O(1)$的转移。

#include<bits/stdc++.h>

using namespace std;

#define int long long

using i32 = int32_t;
using vi = vector<int>;

const int inf = 1e18;
const int mod = 998244353;

int power(int x, int y) {
    int res = 1;
    while (y) {
        if (y & 1) res = res * x % mod;
        x = x * x % mod, y /= 2;
    }
    return res;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n;
    cin >> n;
    vi a(n);
    for (auto &i: a) cin >> i;
    vi f(n + 1);
    int res = 0, invN = inv(n);
    for (int i = n - 1; i >= 0; i--)
        res = (res + res * invN % mod + a[i]) % mod;
    cout << res * invN % mod << "\n";
    return 0;
}