排列组合：构造RGB序列https://atcoder.jp/contests/abc266/tasks/abc266_g

输入 R,G,B,K，要求构造字符串（只由 R,G,B 构成），满足：R 出现 R 次，G 出现 G 次，B 出现 B 次，RG 出现 K 次。问可以构造出多少种这样的串。

首先考虑 G,B 的分配（不会影响到 RG），方案数为\(C_{G+B}^G\)；
然后在排好的 GB 串中选 K 个 G，插入 R, 变成 RG，方案数为 \(C_G^K\)；
最后还剩 R−K 个 R，放进去不能产生新的 RG，所以只能填在 B 前或 R 前，方案数为\(C_{R-K+B+K}^{R-K}=C_{R+B}^{R-K}\)

故答案为 \(C_{G+B}^G *C_G^K *C_{R+B}^{R-K}\)

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;
const int P = 998244353, N = 3e6 + 5;
int R, G, B, K;

using i64 = long long;
// assume -P <= x < 2P
int norm(int x) {
    if (x < 0) {
        x += P;
    }
    if (x >= P) {
        x -= P;
    }
    return x;
}
template<class T>
T power(T a, i64 b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

struct Z {
    int x;
    Z(int x = 0) : x(norm(x)) {}
    Z(i64 x) : x(norm(x % P)) {}
    int val() const {
        return x;
    }
    Z operator-() const {
        return Z(norm(P - x));
    }
    Z inv() const {
        assert(x != 0);
        return power(*this, P - 2);
    }
    Z &operator*=(const Z &rhs) {
        x = i64(x) * rhs.x % P;
        return *this;
    }
    Z &operator+=(const Z &rhs) {
        x = norm(x + rhs.x);
        return *this;
    }
    Z &operator-=(const Z &rhs) {
        x = norm(x - rhs.x);
        return *this;
    }
    Z &operator/=(const Z &rhs) {
        return *this *= rhs.inv();
    }
    friend Z operator*(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res *= rhs;
        return res;
    }
    friend Z operator+(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res += rhs;
        return res;
    }
    friend Z operator-(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res -= rhs;
        return res;
    }
    friend Z operator/(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res /= rhs;
        return res;
    }
    friend std::istream &operator>>(std::istream &is, Z &a) {
        i64 v;
        is >> v;
        a = Z(v);
        return is;
    }
    friend std::ostream &operator<<(std::ostream &os, const Z &a) {
        return os << a.val();
    }
};

Z fac[N], inv[N];
 
Z C(int x,int y){
	return fac[x] * inv[y] * inv[x-y];
}

signed main () {
    cin >> R >> G >> B >> K;
    int N = R + G + B;

    fac[0] = 1;
    for (int i = 1; i <= N; i++)    fac[i] = fac[i-1] * i;
    inv[N] = fac[N].inv();
    for (int i = N; i; i--)     inv[i-1] = inv[i] * i; 

    Z ans = C(G+B, G) * C(G, K) * C(B+R, R-K);
    cout << ans;
}