A 考察组合数学 dp
f[i]为前i个人的组成方案，计算时用组合数学方法计算

# include <bits/stdc++.h>
using namespace std;
#define int long long
typedef long long LL;
const int mod=998244353,N=1e6+10;
int fac[N],ifac[N],f[N];
LL C(int n, int m)
{
    return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
LL fpow(LL a, LL b)
{
    LL res = 1;
    while(b)
    {
        if(b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
void init(int n)
{
fac[0]=1;
for(int i=1;i<=n;i++)
{
fac[i]=fac[i-1]*i%mod;
}
ifac[n] = fpow(fac[n], mod - 2);
for(int i=n;i>=1;i--)
{
ifac[i-1]=ifac[i]*i%mod;
}
    f[0]=1;
    f[1]=0;
    f[2]=1;
    f[3]=1;
    for(int i=4;i<=n;i++)
    {
    f[i]=C(i-1,1)*f[i-2]%mod;
    f[i]+=C(i-1,2)*f[i-3]%mod;
    f[i]+=C(i-1,3)*f[i-4]%mod;
    }
}
void slove(int n) 
{
cout<<f[n]%mod<<endl;
}
signed main()
{
int T;
cin>>T;
init(1e6);
//cout<<ifac[1000000]<<endl;
while(T--)
{
    int x;
    cin>>x;
slove(x);
}
return 0;
}

C
考虑预先处理出1e6内所有数约数。用约数筛处理

# include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
int n;
vector<int> pri;
bool not_prime[N];
int d[N], num[N],s[N];
void pre(int n)
{
  d[1] = 1;
  for (int i = 2; i <= n; ++i)
  {
    if (!not_prime[i])
    {
      pri.push_back(i);
      d[i] = 2;
      num[i] = 1;
    }
    for (int pri_j : pri)
    {
      if (i * pri_j > n) break;
      not_prime[i * pri_j] = true;
      if (i % pri_j == 0)
      {
        num[i * pri_j] = num[i] + 1;
        d[i * pri_j] = d[i] / num[i * pri_j] * (num[i * pri_j] + 1);
        break;
      }
      num[i * pri_j] = 1;
      d[i * pri_j] = d[i] * 2;
    }
  }
}
signed main()
{
        pre(1e6+1);
        for(int i=1;i<=1e6;i++)
        {
        s[i]=s[i-1]+(d[i]<=4);
        }
        int T;
        cin>>T;
        while(T--)
        {
                int l,r;
                cin>>l>>r;
                cout<<s[r]-s[l-1]<<endl;
        }
}

B
发现反对称矩阵对称位置和为零，需要对称位置之和是偶数；

# include <bits/stdc++.h>
using namespace std;
int n;
int main()
{
    cin>>n;
    vector a(n+1,vector<int>(n+1));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            cin>>a[i][j];
        }
    }
    bool st=1;
    vector b(n + 1, vector<int>(n + 1));
    vector c(n + 1, vector<int>(n + 1));
    for(int i=1;i<=n;i++)
    {
        for(int j=i;j<=n;j++)
        {
        if(i!=j)
        {
        int val=a[i][j]+a[j][i];
            if(val&1)
            {
            st=0;
            cout<<"NO";
            return 0;
            }
            else 
            {
            int x = val / 2;
            b[i][j] = b[j][i] = x;
            int y = a[i][j] - x;
            c[i][j] = y;
            c[j][i] = -y;
            }
        }
        else
        {
        b[i][j] = a[i][j];
        }
        }
    }
    if(st)
    {
    cout<<"YES"<<endl;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
        cout<<b[i][j]<<" ";
        }
        cout<<endl;
    }
        for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
        cout<<c[i][j]<<" ";
        }
        cout<<endl;
    }
    }
    return 0;
}