CodeForces 1065E. Side Transmutations 计数

昨天不该早点走的....

 

首先操作限制实际上是一个回文限制

每个$b[i] - b[i - 1]$互不干扰，不妨设这个串关于中心点对称的这么一对区间的串分别为$(S_1, S_2)$

题目的限制相当与存在$(T_1, T_2)$使得$T_1 = inv(S_2) \;and\;T_2 = inv(S_1)$

考虑一对串$(S_1, S_2)$被计数多少次，我们分类讨论一下

 

一个长为$L$的子串的方案数为$S^L$，即为$f(L)$

一个长为$L$,字符集为$S$的区间，形成回文串的方案数为$S^{\frac{L +1}{2}}$，记为$g(L)$

如果$(S_1, S_2)$中有两个回文串，会被算重0次，否则都会被算重1次

那么方案数为$(f(L)^2 - g(L)^2) / 2 + g(L) * g(L)$

化简一下，$f(L) * (f(L) + 1) / 2$

 

复杂度$O(n \log n)$

#include 
#include 
#include 
#include 
#include 
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair
    #define mp make_pair
    #define pb push_back
    #define tpr template 
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)    
    extern inline char gc() {
        static char RR[23456], *S = RR + 23333, *T = RR + 23333;
        if(S == T) fread(RR, 1, 23333, stdin), S = RR;
        return *S ++;
    }
    inline int read() {
        int p = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } 
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
using namespace std;
using namespace remoon;

#define mod 998244353
#define iv2 499122177
#define sid 200050

inline int fp(int a, int k) {
    int ret = 1;
    for( ; k; k >>= 1, a = 1ll * a * a % mod)
        if(k & 1) ret = 1ll * ret * a % mod;
    return ret;
}

int n, m, S;
int b[sid];

int main() {
    n = read(); m = read(); S = read();
    rep(i, 1, m) b[i] = read();    
    int ans = fp(S, n - (b[m] * 2));
    rep(i, 1, m) {
        int L = b[i] - b[i - 1];
        ans = 1ll * ans * fp(S, L)  % mod * (fp(S, L) + 1)  % mod * iv2 % mod;
    }
    write(ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/reverymoon/p/9779929.html