Niu Mei's math problems

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
#define int long long
const int N = 1e7 + 5;
const int mod = 998244353;
int n, k;

int fac[N], infac[N];
int a[N];
int pow2[N];
int qmi (int a, int b, int c) {
    int ans = 1 %c;
    while (b) {
        if (b & 1) ans =ans * a % c;
        b >>= 1;
        a = a * a % c;
    } 
    return (ans %c+ c)%c;
}
int C(int n,int m)
{
    if(n<m) return 0;
    return fac[n]*infac[m]%mod*infac[n-m]%mod;
}
void init () {
	//cout << qmi(3, mod- 2, mod) << endl;
    pow2[0] = 1;
    fac[0] = 1;
    for (int i =1; i< N ; i++)
        fac[i] = fac[i - 1] * i %mod;
    for (int i = 1; i < N; i++)
        pow2[i] = pow2[i - 1] * 2 %mod;
        
    infac[N - 1] = qmi (fac[N - 1], mod - 2, mod) %mod;
    infac[0] = infac[1] = 1;
    for (int i = N - 1; i>= 1;  i--)
    	infac[i - 1] = infac[i] * i %mod;
    
}
void solve() {
	 
    cin >> n >> k;
     init();
   // cout << C(3, 1) << endl; 
	
    int one = 0, two = 0;
    
    for (int i = 1; i <= n; i ++ )
    {
        cin >> a[i];
        if (a[i] == 1) one ++;
        else if (a[i] == 2) two ++;
    }
   
 
   int ans = 0;
    for (int i = 0; i <= k; i++) {
        ans = ((ans + C(one, i) * C(two, k - i) %mod * pow2[k - i] %mod + mod) %mod + mod) %mod;
    }
    
    cout << ans << endl;
}
signed main () {
    int t;
    t = 1;
    while (t --) solve();
}

By looking at the data range , Statistics 1 and 2 The number of ,0 No contribution , Sure . This continuous summation can be regarded as a sequence . And the solution of inverse element , The linear method can be found by observing the formula , At first, I didn't think from the perspective of formula