A：S 老师的公式

签到题

using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 10086;
map<char, int> mp;
int n, m;

signed main() {
    cin >> n;
    int num1 = (1 + n) * n / 2;
    int sum = 1;
    for (int i = 1; i <= n; i++) {
        sum *= i;
        sum %= num1;
    }
    cout << gcd(num1, sum) << endl;
    return 0;
}

B：S 老师的签到

需要用广度搜索加标记数组去减少空间使用量

using namespace std;
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 10086;
int n, m;
char a[1500][1500];
string ans;
vector<pair<int,int>>ve;
int vis[1500][1500];
void bfs(){
    queue<pair<int,int>>q;
    q.push({0, 0});
    while(!q.empty()){
        int num=q.size();
        char s='z';
        int f=0;
        for(int i=0;i<num;i++){
            pair op=q.front();
            q.pop();
            int x=op.first;
            int y=op.second;

            if(vis[x][y])continue;
            vis[x][y]=1;
            s=min(a[x][y],s);
            if(x==n-1&&y==m-1) {
                f=1;
                break;
            }
            if(x+1<n&&y+1<m){
                if(a[x+1][y]==a[x][y+1]){
                    ve.push_back({x+1,y});
                    ve.push_back({x , y+1});
                }else if(a[x+1][y]<a[x][y+1]){
                    ve.clear();
                    ve.push_back({x+1,y});
                }else{
                    ve.clear();
                    ve.push_back({x,y+1});
                }
                continue;
            }
            if(x+1<n){
                ve.push_back({x+1,y});
            }
            if(y+1<m){
                ve.push_back({x , y+1});
            }
        }
        ans+=s;
        if(f==1)return ;
        for(auto it:ve){
            q.push({it.first,it.second});
        }
    }
}
signed main() {
    cin>>n>>m;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            cin>>a[i][j];
        }
    }
    bfs();
    cout<<ans<<endl;
    return 0;
}

C：S 老师的求和

数学公式推导加逆元

using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 998244353;
int n, m;
int a,b,x;
int ksm(int X,int y){
    int res=1;
    while(y){
        if(y&1)res=res*X%mod;
        X=X*X%mod;
        y>>=1;
    }
    return res%mod;
}
int L1(int A,int B,int X){
    int op=A*X%mod;
    return (op+B)%mod;
}
int L2(int A,int B,int X){
    int op=X*B%mod;
    int num=(A*(1+X)%mod*X%mod*ksm(2,mod-2))%mod;
    return (op+num)%mod;
}
int L3(int A,int B,int X){
    int op=((1+X)*X%mod*B%mod*ksm(2,mod-2))%mod;
    int num=(A*X%mod*(X+1)%mod*(X+2)%mod*ksm(6,mod-2))%mod;
    return (op+num)%mod;
}
int L4(int A,int B,int X){
    int op=(B*X%mod*(X+1)%mod*(X+2)%mod*ksm(6,mod-2))%mod;
    int num=(A*X%mod*(X+1)%mod*(X+2)%mod*(X+3)%mod*ksm(24,mod-2))%mod;
    return (op+num)%mod;
}
signed main() {
    int _;
    cin>>_;
    while(_--){
        cin>>a>>b>>x;
        cout<<L1(a,b,x)<<" "<<L2(a,b,x)<<" "<<L3(a,b,x)<<" "<<L4(a,b,x)<<endl;
    }
    return 0;
}