【LOJ6485】LJJ 学二项式定理

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【题目链接】

• 点击打开链接

【思路要点】

• $\sum_{t=0}^{3}a_t\sum_{i=0}^{N}\binom{N}{i}s^i[i\%4=t]$
  $=\sum_{t=0}^{3}a_t\sum_{i=0}^{N}\binom{N}{i}s^i\sum_{j=0}^{3}w^{j(i-t)}_4$
  $=\sum_{t=0}^{3}a_t\sum_{j=0}^{3}w^{-jt}_4\sum_{i=0}^{N}\binom{N}{i}s^i w^{ij}_4$
  $=\sum_{t=0}^{3}a_t\sum_{j=0}^{3}w^{-jt}_4(sw^{j}_4+1)^N$
• 时间复杂度 $O(TLogN)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int P = 998244353;
const int w[4] = {1, 911660635, 998244352, 86583718};
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int power(int x, ll y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int main() {
	int T; read(T);
	int inv4 = power(4, P - 2);
	while (T--) {
		ll n; int s, ans = 0;
		read(n), read(s);
		for (int i = 0; i <= 3; i++) {
			int x; read(x);
			for (int j = 0; j <= 3; j++)
				update(ans, 1ll * x * w[(4 - i) * j % 4] % P * power((1ll * s * w[j] + 1) % P, n) % P);
		}
		writeln(1ll * ans * inv4 % P);
	}
	return 0;
}