AtCoderARC099 F - Eating Symbols Hard

Posted 2020-11-20 ivorysi

题解

一道神奇的题

我们把操作S构成的A数组用一个多项式表示出来
(t(S) = sum_{i = -10^9}^{10^9} A_{i}X^{i})
如果往S前面添加一个字符的话
(t<(S) = t(S)X^{-1})
(t>(S) = t(S)X)
(t+(S) = t(S) + 1)
(t-(S) = t(S) - 1)
那么我们对于最终的序列求一个哈希值c，如果一段区间操作后的结果和c一样的话就有
(t_{S_i}t_{S_{i + 1}}...t_{S_j}(0) = c)
由于这些操作可逆，可以一层一层拆开
可以得到
(t_{S_N}^{-1}...t_{S_i}^{-1}t_{S_i}t_{S_{i + 1}}...t_{S_j}(0) = t_{S_N}^{-1}...t_{S_i}^{-1}(c))
那么我们可以得到
(t_{S_N}^{-1}...t_{S_{j + 1}}^{-1} (0) = t_{S_N}^{-1}...t_{S_{i}}^{-1}(c))
这个后缀积可以线性处理出来，维护未知数前的系数即可

然后就是愉快的用map查询了

那么，冲突怎么考虑？题解说是冲突的概率在2N/模数大小，让用6个，然而我写的不优美，T掉了，改成5个卡着时限A了,感觉用不上太多也是对的啊

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <map>
#define enter putchar(‘n‘)
#define space putchar(‘ ‘)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pii pair<int,int>
#define eps 1e-7
#define MAXN 250005
#define MOD 999999137
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
typedef vector<int> poly;
 
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) {putchar(‘-‘);x = -x;}
    if(x >= 10) {
        out(x / 10);
    }
    putchar(‘0‘ + x % 10);
}
int B[] = {0,823,727,401,271,571};
int InvB[10];
int f[8][MAXN],g[8][MAXN],N,h[8][MAXN];
char s[MAXN];
map<int,int> MK[8];
 
 
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
void update(int &x,char c,int id) {
    if(c == ‘<‘) x = mul(x,InvB[id]);
    else if(c == ‘>‘) x = mul(x,B[id]);
    else if(c == ‘+‘) x = inc(x,1);
    else x = inc(x,MOD - 1);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
void Solve() {
    for(int i = 1 ; i <= 5 ; ++i) InvB[i] = fpow(B[i],MOD - 2);
    read(N);
    scanf("%s",s + 1);
    int c[10] = {0};
    for(int k = 1 ; k <= 5 ; ++k) {
        for(int i = N ; i >= 1 ; --i) {
            update(c[k],s[i],k);
        }
    }
    for(int k = 1 ; k <= 5 ; ++k) h[k][N + 1] = 1,g[k][N + 1] = c[k],MK[k][0] += 1;
    int64 ans = 0;
    for(int i = N ; i >= 1 ; --i) {
        int add = N - i + 1;
        for(int k = 1 ; k <= 5 ; ++k) {
            if(s[i] == ‘<‘) {
                g[k][i] = inc(g[k][i + 1],MOD - mul(h[k][i + 1],c[k]));
                h[k][i] = mul(h[k][i + 1],B[k]);
                g[k][i] = inc(g[k][i],mul(h[k][i],c[k]));
                f[k][i] = f[k][i + 1];      
            }
            else if(s[i] == ‘>‘) {
                g[k][i] = inc(g[k][i + 1],MOD - mul(h[k][i + 1],c[k]));
                h[k][i] = mul(h[k][i + 1],InvB[k]);
                g[k][i] = inc(g[k][i],mul(h[k][i],c[k]));
                f[k][i] = f[k][i + 1];
            }
            else if(s[i] == ‘+‘) {
                h[k][i] = h[k][i + 1];
                g[k][i] = inc(g[k][i + 1],MOD - h[k][i]);
                f[k][i] = inc(f[k][i + 1],MOD - h[k][i]);
            }
            else {
                h[k][i] = h[k][i + 1];
                g[k][i] = inc(g[k][i + 1],h[k][i]);
                f[k][i] = inc(f[k][i + 1],h[k][i]);
            }
            add = min(add,MK[k][g[k][i]]);
            MK[k][f[k][i]] += 1;
        }
        ans += add;
    }
    out(ans);enter; 
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}