标签：目的 and 计数 namespace 复杂 include make 分类 log
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$S^{\frac{L}{2}}$，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$f(L) * f(L) / 2 - g(L) * g(L)$
化简一下，$f(L) * (f(L) + 1) / 2$
复杂度$O(n \log n)$
#include <vector> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> 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<int, int> #define mp make_pair #define pb push_back #define tpr template <typename ra> #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; }
CodeForces 1065E. Side Transmutations 计数
标签：目的 and 计数 namespace 复杂 include make 分类 log
原文地址：https://www.cnblogs.com/reverymoon/p/9779929.html
