[BZOJ 3160] 万径人踪灭

3160: 万径人踪灭

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 2662  Solved: 1447
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Description

Input

Output

 

Sample Input

 

Sample Output

 

HINT

 

考虑用总回文序列数减去回文串数

对于每个对称轴，只需要知道有多少对关于它对称然后计算

先把$a$置为$1$，$b$置为$0$做一次卷积，然后$a$置为$0$，$b$置为$1$做一次卷积

两次加起来再稍微处理一下就行了

回文串数用马拉车求

#include <bits/stdc++.h>
using namespace std;
const int maxn = 262144, mod = 1e9 + 7;
const double PI = acos(-1.0);

struct comp{
    double x, y;
    comp(double _x = 0, double _y = 0){
        x = _x;
        y = _y;
    }
    friend comp operator * (const comp &a, const comp &b){
        return comp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
    }
    friend comp operator + (const comp &a, const comp &b){
        return comp(a.x + b.x, a.y + b.y);
    }
    friend comp operator - (const comp &a, const comp &b){
        return comp(a.x - b.x, a.y - b.y);
    }
}f[maxn], g[maxn];
int rev[maxn];
void dft(comp A[], int len, int kind){
    for(int i = 0; i < len; i++){
        if(i < rev[i]){
            swap(A[i], A[rev[i]]);
        }
    }
    comp wn, s, t, tmp;
    for(int i = 1; i < len; i <<= 1){
        wn = comp(cos(PI / i), kind * sin(PI / i));
        for(int j = 0; j < len; j += (i << 1)){
            tmp = comp(1, 0);
            for(int k = 0; k < i; k++){
                s = A[j + k], t = tmp * A[i + j + k];
                A[j + k] = s + t;
                A[i + j + k] = s - t;
                tmp = tmp * wn;
            }
        }
    }
    if(kind == -1) for(int i = 0; i < len; i++) A[i].x /= len;
}
void init(int &len, int n){
    int L = 0;
    for(len = 1; len < n + n - 1; len <<= 1, L++);
    for(int i = 0; i < len; i++){
        rev[i] = rev[i >> 1] >> 1 | (i & 1) << L - 1;
    }
}
int bin[100000 + 10];
char str[200000 + 10];
int slen, p[200000 + 10] = {0};
int main(){
    scanf("%s", str);
    int n = strlen(str), len;
    init(len, n);
    for(int i = 0; i < n; i++){
        if(str[i] == 'a') f[i].x = 1;
    }
    for(int i = 0; i < n; i++){
        if(str[i] == 'b') g[i].x = 1;
    }
    dft(f, len, 1);
    dft(g, len, 1);
    for(int i = 0; i < len; i++) g[i] = g[i] * g[i] + f[i] * f[i];
    dft(g, len, -1);
    for(int i = 0; i < n; i++) g[i << 1].x++;
    bin[0] = 1;
    for(int i = 1; i <= 100000; i++){
        bin[i] = bin[i - 1] << 1;
        if(bin[i] >= mod) bin[i] -= mod; 
    }
    int ans = 0;
    for(int i = 0; i < len; i++){
        ans += bin[(int)(g[i].x + 0.5) >> 1] - 1;
        if(ans >= mod) ans -= mod; 
    }    
    for(int i = n - 1; ~i; i--){
        str[i * 2 + 2] = str[i];
        str[i * 2 + 3] = '$';
    }
    str[0] = '~'; str[1] = '$';
    slen = 2 * n + 2;
    int mr = 0, mid;
    for(int i = 1; i < slen; i++){
        if(i <= mr){
            p[i] = min(mr - i + 1, p[(mid << 1) - i]);
        }
        while(str[i + p[i]] == str[i - p[i]]) p[i]++;
        if(p[i] + i > mr) mr = p[i] + i - 1, mid = i;
        ans -= p[i] >> 1;
        if(ans < 0) ans += mod;
    }
    printf("%d\n", ans);
    return 0;
} 

 

来源：https://www.cnblogs.com/ruoruoruo/p/11787263.html