【洛谷】P4199 万径人踪灭

题解

每种字符跑一遍FFT，得到\(i + j = k\)时匹配的个数（要÷2，对于相同位置的最后再加上

然后算出\(2^{cnt[k]}\)的和，最后再减去用mancher匹配出的连续回文子串的个数即可

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353,MAXL = (1 << 20);
const int M = 1000000007;

int pw[400005],f[400005],N,W[MAXL + 5],ans,r[400005],mx,pos,cnt[400005];
char s[400005];
char t[400005];
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = 1LL * res * t % MOD;
	t = 1LL * t * t % MOD;
	c >>= 1;
    }
    return res;
}
void NTT(int *p,int L,int on) {
    for(int i = 1 , j = L >> 1 ; i < L - 1 ; ++i) {
	if(i < j) swap(p[i],p[j]);
	int k = L >> 1;
	while(j >= k) {
	    j -= k;
	    k >>= 1;
	}
	j += k;
    }
    for(int h = 2 ; h <= L ; h <<= 1) {
	int wn = W[(MAXL + MAXL * on / h) % MAXL];
	for(int k = 0 ; k < L ; k += h) {
	    int w = 1;
	    for(int j = k ; j < k + h / 2 ; ++j) {
		int u = p[j],t = 1LL * w * p[j + h / 2] % MOD;
		p[j] = u + t >= MOD ? u + t - MOD : u + t;
		p[j + h / 2] = u - t >= 0 ? u - t : u - t + MOD;
		w = 1LL * w * wn % MOD;
	    }
	}
    }
    if(on == -1) {
	int InvL = fpow(L,MOD - 2);
	for(int i = 0 ; i < L ; ++i) p[i] = 1LL * InvL * p[i] % MOD;
    }
}
void Init() {
    W[0] = 1;
    W[1] = fpow(3,(MOD - 1) / MAXL);
    for(int i = 2; i < MAXL ; ++i) W[i] = 1LL * W[i - 1] * W[1] % MOD;
    pw[0] = 1;
    for(int i = 1 ; i <= 400000 ; ++i) pw[i] = 1LL * pw[i - 1] * 2 % M;
}
void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i <= N ; ++i) {
	if(s[i] == 'a') f[i] = 1;
    }
    int len = 1;
    while(len <= 2 * N) {len <<= 1;}
    NTT(f,len,1);
    for(int i = 0 ; i < len ; ++i) f[i] = 1LL * f[i] * f[i] % MOD;
    NTT(f,len,-1);
    for(int i = 0 ; i < len ; ++i) {
	cnt[i] += f[i] / 2;
    }
    memset(f,0,sizeof(f));
    for(int i = 1 ; i <= N ; ++i) {
	if(s[i] == 'b') f[i] = 1;
    }
    NTT(f,len,1);
    for(int i = 0 ; i < len ; ++i) f[i] = 1LL * f[i] * f[i] % MOD;
    NTT(f,len,-1);
    for(int i = 0 ; i < len ; ++i) {
	cnt[i] += f[i] / 2;
    }
    for(int i = 0 ; i < len ; ++i) {
	if(i % 2 == 0 && i / 2 >= 1 && i / 2 <= N) ++cnt[i];
	ans = ans + pw[cnt[i]] - 1 >= M ? ans + pw[cnt[i]] - 1 - M : ans + pw[cnt[i]] - 1;
    }
    int p = 1;
    t[++p] = '$';
    for(int i = 1 ; i <= N ; ++i) {
	t[++p] = s[i];
	t[++p] = '$';
    }
    mx = 1,pos = 1,r[1] = 1;
    for(int i = 2 ; i <= p ; ++i) {
	if(mx >= i) r[i] = min(r[2 * pos - i],mx - i + 1);
	else r[i] = 1;
	while(i + r[i] <= p && i - r[i] >= 1 && t[i + r[i]] == t[i - r[i]]) ++r[i];
	if(i + r[i] - 1 > mx) {
	    mx = i + r[i] - 1;
	    pos = i;
	}
    }
    for(int i = 1 ; i <= p ; ++i) {
	r[i] /= 2;
	ans = ans - r[i] >= 0 ? ans - r[i] : ans - r[i] + M;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}