利用杨辉三角

递推模式

#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998'244'353;
long long n, k, C[1005][1005];
int main()
{
    cin >> n >> k;
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= i; j++)
            if (j == 0 || j == i)
                C[i][j] = 1;
            else
                C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
    cout << C[n][k] << "\n";
    return 0;
}

递归模式

#include <bits/stdc++.h>
using namespace std;
int C(int x, int y);
int n, m;
const int mod = 998244353;
int a[1005][1005]; //定义在全局自动初始化为全0了
int C(int x, int y)
{
    if (a[x][y] != 0)
        return a[x][y];
    if (y == 0 || y == x)
        return a[x][y] = 1;
    return a[x][y] = (C(x - 1, y) + C(x - 1, y - 1)) % mod;
}
int main()
{
    cin >> n >> m;
    cout << C(n, m);
    return 0;
}

除法用逆元处理

#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998'244'353;
long long inv[1005];
long long n, k, ans;
int main()
{
    inv[1] = 1;
    for (int i = 2; i <= 1000; i++)
        inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;

    cin >> n >> k;
    ans = 1;
    for (int i = n; i >= n - k + 1; i--)
        ans = (ans * i) % MOD;
    for (int i = 1; i <= k; i++)
        ans = (ans * inv[i]) % MOD;
    cout << ans << "\n";
    return 0;
}