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Niu Mei's math problems
2022-07-05 06:16:00 【whitewall_ nine】
#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
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