【省内训练2019-06-06】计数_省内训练2019-06-28-优快云博客

【思路要点】

考虑枚举一对颜色相等的位置 $i, j$ ，计算它们的贡献，则有
$Ans=∑i=1N∑j=i+1N[ci=cj]2j−i−1∑k=0min{i−1,N−j}(i−1k)×(N−jk)Ans=\sum_{i=1}^{N}\sum_{j=i+1}^{N}[c_i=c_j]2^{j-i-1}\sum_{k=0}^{min\{i-1,N-j\}}\binom{i-1}{k}\times\binom{N-j}{k}$
注意到 $∑i=0min{a,b}(ai)×(bi)\sum_{i=0}^{min\{a,b\}}\binom{a}{i}\times\binom{b}{i}$ 实际上表示在两堆物品中选择相等个的方案数，而 $x+1)^p(x^{-1}+1)^q$ 第 $x^i$ 的系数应当为 $(p+qi+q)\binom{p+q}{i+q}$ ，因此 $∑i=0min{a,b}(ai)×(bi)=(a+ba)\sum_{i=0}^{min\{a,b\}}\binom{a}{i}\times\binom{b}{i}=\binom{a+b}{a}$
从而
$Ans=∑i=1N∑j=i+1N[ci=cj]2j−i−1(N−j+i−1N−j)Ans=\sum_{i=1}^{N}\sum_{j=i+1}^{N}[c_i=c_j]2^{j-i-1}\binom{N-j+i-1}{N-j}$
直接 $N T T$ 优化上式即可。
时间复杂度 $O (N L o g N)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 524288;
const int P = 998244353;
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("");
}
namespace NTT {
	const int MAXN = 524288;
	const int P = 998244353;
	const int G = 3;
	int power(int x, int 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;
	}
	int N, Log, home[MAXN];
	void NTTinit() {
		for (int i = 0; i < N; i++) {
			int ans = 0, tmp = i;
			for (int j = 1; j <= Log; j++) {
				ans <<= 1;
				ans += tmp & 1;
				tmp >>= 1;
			}
			home[i] = ans;
		}
	}
	void NTT(int *a, int mode) {
		for (int i = 0; i < N; i++)
			if (home[i] < i) swap(a[i], a[home[i]]);
		for (int len = 2; len <= N; len <<= 1) {
			int delta;
			if (mode == 1) delta = power(G, (P - 1) / len);
			else delta = power(G, P - 1 - (P - 1) / len);
			for (int i = 0; i < N; i += len) {
				int now = 1;
				for (int j = i, k = i + len / 2; k < i + len; j++, k++) {
					int tmp = a[j];
					int tnp = 1ll * a[k] * now % P;
					a[j] = (tmp + tnp) % P;
					a[k] = (tmp - tnp + P) % P;
					now = 1ll * now * delta % P;
				}
			}
		}
		if (mode == -1) {
			int inv = power(N, P - 2);
			for (int i = 0; i < N; i++)
				a[i] = 1ll * a[i] * inv % P;
		}
	}
	void times(int *a, int *b, int *c, int limit) {
		N = 1, Log = 0;
		while (N < 2 * limit) {
			N <<= 1;
			Log++;
		}
		for (int i = limit; i < N; i++)
			a[i] = b[i] = 0;
		NTTinit();
		NTT(a, 1);
		NTT(b, 1);
		for (int i = 0; i < N; i++)
			c[i] = 1ll * a[i] * b[i] % P;
		NTT(c, -1);
	}
}
int fac[MAXN], inv[MAXN];
int power(int x, int 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;
}
int binom(int x, int y) {
	if (y > x) return 0;
	else return 1ll * fac[x] * inv[y] % P * inv[x - y] % P;
}
void init(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = 1ll * fac[i - 1] * i % P;
	inv[n] = power(fac[n], P - 2);
	for (int i = n - 1; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1ll) % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
char s[MAXN];
int n, ans, two[MAXN], a[MAXN], b[MAXN], c[MAXN];
int main() {
	scanf("%s", s + 1);
	init(n = strlen(s + 1)), two[0] = 1;
	for (int i = 1; i <= n; i++) {
		if (s[i] == '0') a[i - 1] = inv[i - 1], b[n - i] = inv[n - i];
		else a[i - 1] = b[n - i] = 0;
		two[i] = two[i - 1] * 2 % P;
	}
	NTT :: times(a, b, c, n);
	for (int i = 0; i <= n - 2; i++)
		update(ans, 1ll * c[i] * fac[i] % P * two[n - 2 - i] % P);
	for (int i = 1; i <= n; i++)
		if (s[i] == '1') a[i - 1] = inv[i - 1], b[n - i] = inv[n - i];
		else a[i - 1] = b[n - i] = 0;
	NTT :: times(a, b, c, n);
	for (int i = 0; i <= n - 2; i++)
		update(ans, 1ll * c[i] * fac[i] % P * two[n - 2 - i] % P);
	writeln(ans);
	return 0;
}