BZOJ5372: PKUSC2018神仙的游戏

传送门

Sol

自己还是太 (naive) 了，上来就构造多项式和通配符直接匹配，然后遇到 (border) 相交的时候就 (gg) 了
神仙的游戏蒟蒻还是玩不来
一个小小的性质：
存在长度为 (len) 的 (border) 的充要条件是 (forall i,s_i=s_{n-len+i})
等价于按照 (n-len) 的剩余系分类，那么每一类都要求不同时含有 (0,1)
考虑两个位置 (i,j) 分别为 (0,1) 会对于哪一些长度的 (border) 有影响
显然是满足 (|i-j|equiv 0 (mod~n-len)) 的 (len)，即 ((n-len)|(|i-j|))
设 (f_x) 表示 (|i-j|=x) 是否存在 (s_i,s_j) 分别为 (0,1)
这个是一个经典套路
只要对于 (s_i=1) 和 (s_i=0) 分别构造函数，(FFT) 一下就好了

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

const int maxn(4e6 + 5);
const double pi(acos(-1));

struct Complex {
	double a, b;

	inline Complex() {
		a = b = 0;
	}

	inline Complex(double x, double y) {
		a = x, b = y;
	}

	inline Complex operator +(Complex x) const {
		return Complex(a + x.a, b + x.b);
	}
    
	inline Complex operator -(Complex x) const {
		return Complex(a - x.a, b - x.b);
	}

	inline Complex operator *(Complex x) const {
		return Complex(a * x.a - b * x.b, a * x.b + b * x.a);
	}
} a[maxn], b[maxn], w[maxn];

int deg, r[maxn], l;

inline void Init(int n) {
	register int i, k;
	for (deg = 1, l = 0; deg < n; deg <<= 1) ++l;
	for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for (i = 1; i < deg; i <<= 1)
		for (k = 0; k < i; ++k) w[deg / i * k] = Complex(cos(pi / i * k), sin(pi / i * k));
}

inline void FFT(Complex *p, int opt) {
	register int i, j, k, t;
	register Complex wn, x, y;
	for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
	for (i = 1; i < deg; i <<= 1)
		for(t = i << 1, j = 0; j < deg; j += t)
			for (k = 0; k < i; ++k) {
				wn = w[deg / i * k];
				if (opt == -1) wn.b *= -1;
				x = p[j + k], y = wn * p[i + j + k];
				p[j + k] = x + y, p[i + j + k] = x - y;
			}
}

int n, len, f[maxn], g[maxn];
ll ans;
char s[maxn];

int main() {
	register int i, j;
	scanf(" %s", s + 1), n = strlen(s + 1);
	for (i = 1; i <= n; ++i) f[i] = s[i] == '0', g[i] = s[n - i + 1] == '1';
	for (len = 1; len <= n + n; len <<= 1);
	for (i = 1; i <= n; ++i) a[i].a = f[i], b[i].a = g[i];
	Init(len), FFT(a, 1), FFT(b, 1);
	for (i = 0; i < len; ++i) a[i] = a[i] * b[i];
	FFT(a, -1);
	for (i = 0; i <= n + n; ++i) f[i] = (int)(a[i].a / len + 0.5);
	for (i = 1; i <= n; ++i) g[i] = f[n + 1 - i] + f[n + 1 + i];
	for (i = 1; i <= n; ++i)
		for (j = i; j <= n; j += i) g[i] |= g[j];
	for (i = 1; i <= n; ++i) if (!g[n - i]) ans ^= 1LL * i * i;
	printf("%lld
", ans);
    return 0;
}