#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define all(x) begin(x), end(x)
#define sz(x) int((x).size())
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
#ifdef LOCAL
auto operator<<(auto& o, auto x) -> decltype(x.first, o);
auto operator<<(auto& o, auto x) -> decltype(x.end(), o) {
o << "{";
for (int i = 0; auto y : x) o << ", " + !i++ * 2 << y;
return o << "}"; }
auto operator<<(auto& o, auto x) -> decltype(x.first, o) {
return o << "(" << x.first << ", " << x.second << ")"; }
void __print(auto... x) { ((cerr << x << " "), ...) << endl; }
#define debug(x...) __print("[" #x "]:", x)
#else
#define debug(...) 2137
#endif
template<int N>
struct IntSet {
static constexpr int B = 64;
uint64_t V[N / B + 1] = {};
IntSet<(N < B + 1 ? 0 : N / B + 1)> up;
bool has(int i) { return (V[i / B] >> i) & 1; }
void add(int i) {
if (!V[i / B]) up.add(i / B);
V[i / B] |= 1ull << i;
}
void del(int i) {
if (!(V[i / B] &= ~(1ull << i))) up.del(i / B);
}
int next(int i) { // j > i such that j inside or -1
auto x = V[i / B] >> i;
if (x &= ~1) return i + __builtin_ctzll(x);
return (i = up.next(i / B)) < 0 ? i :
i * B + __builtin_ctzll(V[i]);
}
int prev(int i) { // j < i such that j inside or -1
auto x = V[i / B] << (B - i - 1);
if (x &= INT64_MAX)
return i-__builtin_clzll(x);
return (i = up.prev(i / B)) < 0 ? i :
i * B + B - 1 - __builtin_clzll(V[i]);
}
};
template<>
struct IntSet<0> {
void add(int) {} void del(int) {}
int next(int) { return -1; }
int prev(int) { return -1; } };
constexpr uint32_t MOD = 998244353;
using M = array<array<uint32_t, 7>, 7>;
const M I = {{
{1, 0, 0, 0, 0, 0, 0},
{0, 1, 0, 0, 0, 0, 0},
{0, 0, 1, 0, 0, 0, 0},
{0, 0, 0, 1, 0, 0, 0},
{0, 0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 0, 1},
}};
// mozliwe opty mnozenia:
// splaszczenie
// zunrollowanie petli (lol)
// optymalniejsze mnozenie (nwm czy znamy optymalne)
struct Tree {
typedef M T;
T f(const T& a, const T& b) {
uint64_t x[7][7] = {};
rep(i, 0, 7) rep(j, 0, 7) rep(k, 0, 7) x[i][k] += (uint64_t)a[i][j] * b[j][k];
T c = {};
rep(i, 0, 7) rep(j, 0, 7) c[i][j] = x[i][j] % MOD;
return c;
}
vector<T> s; int n;
Tree(int N = 0) : s(2*N, I), n(N) {}
void update(int pos, T val) {
for (s[pos += n] = val; pos /= 2;)
s[pos] = f(s[pos * 2], s[pos * 2 + 1]);
}
T query(int b, int e) { // query [b, e)
T ra = I, rb = I;
for (b += n, e += n; b < e; b /= 2, e /= 2) {
if (b % 2) ra = f(ra, s[b++]);
if (e % 2) rb = f(s[--e], rb);
}
return f(ra, rb);
}
};
const int N = 50050;
int n, q;
string s;
M m1[6], m2[6][64];
Tree t1, t2;
IntSet<N> p[6];
uint32_t rozw() {
uint64_t odp = 0;
M a = t1.query(0, n);
rep(i, 0, 6) odp += a[i][6];
debug(odp);
M b = t2.query(0, n);
rep(i, 0, 6) odp += MOD - b[i][6];
return odp % MOD;
}
void napraw(int i) {
int c = s[i] - 'a';
int nxt[6];
rep(j, 0, 6) {
int x = p[j].next(i);
nxt[j] = x != -1 ? x : n;
}
int msk = 0;
rep(j, 0, 6) if (nxt[j] < nxt[c]) msk |= 1 << j;
if (nxt[c] != n) msk |= 1 << c;
t2.update(i, m2[c][msk]);
}
int main() {
cin.tie(0)->sync_with_stdio(0);
rep(i, 0, 6) {
m1[i] = I;
rep(j, 0, 7) m1[i][i][j] = 1;
}
rep(i, 0, 6) rep(msk, 0, 64) {
m2[i][msk] = I;
rep(j, 0, 6) if ((msk >> j) & 1) m2[i][msk][i][j] = 1;
if (!((msk >> i) & 1)) m2[i][msk][i][6] = 1; // +1 jezeli ost wyst
}
cin >> n >> q >> s;
t1 = t2 = Tree(n);
rep(i, 0, n) {
t1.update(i, m1[s[i] - 'a']);
p[s[i] - 'a'].add(i);
}
rep(i, 0, n) napraw(i);
cout << rozw() << '\n';
while (q--) {
int i;
char c;
cin >> i >> c;
i--;
int zm[7];
rep(j, 0, 6) {
zm[j] = p[j].prev(i);
}
zm[6] = i;
p[s[i] - 'a'].del(i);
s[i] = c;
p[c - 'a'].add(i);
for (int x : zm) if (x != -1) napraw(x);
t1.update(i, m1[c - 'a']);
cout << rozw() << '\n';
}
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 | #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i = (a); i < (b); i++) #define all(x) begin(x), end(x) #define sz(x) int((x).size()) using ll = long long; using pii = pair<int, int>; using vi = vector<int>; #ifdef LOCAL auto operator<<(auto& o, auto x) -> decltype(x.first, o); auto operator<<(auto& o, auto x) -> decltype(x.end(), o) { o << "{"; for (int i = 0; auto y : x) o << ", " + !i++ * 2 << y; return o << "}"; } auto operator<<(auto& o, auto x) -> decltype(x.first, o) { return o << "(" << x.first << ", " << x.second << ")"; } void __print(auto... x) { ((cerr << x << " "), ...) << endl; } #define debug(x...) __print("[" #x "]:", x) #else #define debug(...) 2137 #endif template<int N> struct IntSet { static constexpr int B = 64; uint64_t V[N / B + 1] = {}; IntSet<(N < B + 1 ? 0 : N / B + 1)> up; bool has(int i) { return (V[i / B] >> i) & 1; } void add(int i) { if (!V[i / B]) up.add(i / B); V[i / B] |= 1ull << i; } void del(int i) { if (!(V[i / B] &= ~(1ull << i))) up.del(i / B); } int next(int i) { // j > i such that j inside or -1 auto x = V[i / B] >> i; if (x &= ~1) return i + __builtin_ctzll(x); return (i = up.next(i / B)) < 0 ? i : i * B + __builtin_ctzll(V[i]); } int prev(int i) { // j < i such that j inside or -1 auto x = V[i / B] << (B - i - 1); if (x &= INT64_MAX) return i-__builtin_clzll(x); return (i = up.prev(i / B)) < 0 ? i : i * B + B - 1 - __builtin_clzll(V[i]); } }; template<> struct IntSet<0> { void add(int) {} void del(int) {} int next(int) { return -1; } int prev(int) { return -1; } }; constexpr uint32_t MOD = 998244353; using M = array<array<uint32_t, 7>, 7>; const M I = {{ {1, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 1}, }}; // mozliwe opty mnozenia: // splaszczenie // zunrollowanie petli (lol) // optymalniejsze mnozenie (nwm czy znamy optymalne) struct Tree { typedef M T; T f(const T& a, const T& b) { uint64_t x[7][7] = {}; rep(i, 0, 7) rep(j, 0, 7) rep(k, 0, 7) x[i][k] += (uint64_t)a[i][j] * b[j][k]; T c = {}; rep(i, 0, 7) rep(j, 0, 7) c[i][j] = x[i][j] % MOD; return c; } vector<T> s; int n; Tree(int N = 0) : s(2*N, I), n(N) {} void update(int pos, T val) { for (s[pos += n] = val; pos /= 2;) s[pos] = f(s[pos * 2], s[pos * 2 + 1]); } T query(int b, int e) { // query [b, e) T ra = I, rb = I; for (b += n, e += n; b < e; b /= 2, e /= 2) { if (b % 2) ra = f(ra, s[b++]); if (e % 2) rb = f(s[--e], rb); } return f(ra, rb); } }; const int N = 50050; int n, q; string s; M m1[6], m2[6][64]; Tree t1, t2; IntSet<N> p[6]; uint32_t rozw() { uint64_t odp = 0; M a = t1.query(0, n); rep(i, 0, 6) odp += a[i][6]; debug(odp); M b = t2.query(0, n); rep(i, 0, 6) odp += MOD - b[i][6]; return odp % MOD; } void napraw(int i) { int c = s[i] - 'a'; int nxt[6]; rep(j, 0, 6) { int x = p[j].next(i); nxt[j] = x != -1 ? x : n; } int msk = 0; rep(j, 0, 6) if (nxt[j] < nxt[c]) msk |= 1 << j; if (nxt[c] != n) msk |= 1 << c; t2.update(i, m2[c][msk]); } int main() { cin.tie(0)->sync_with_stdio(0); rep(i, 0, 6) { m1[i] = I; rep(j, 0, 7) m1[i][i][j] = 1; } rep(i, 0, 6) rep(msk, 0, 64) { m2[i][msk] = I; rep(j, 0, 6) if ((msk >> j) & 1) m2[i][msk][i][j] = 1; if (!((msk >> i) & 1)) m2[i][msk][i][6] = 1; // +1 jezeli ost wyst } cin >> n >> q >> s; t1 = t2 = Tree(n); rep(i, 0, n) { t1.update(i, m1[s[i] - 'a']); p[s[i] - 'a'].add(i); } rep(i, 0, n) napraw(i); cout << rozw() << '\n'; while (q--) { int i; char c; cin >> i >> c; i--; int zm[7]; rep(j, 0, 6) { zm[j] = p[j].prev(i); } zm[6] = i; p[s[i] - 'a'].del(i); s[i] = c; p[c - 'a'].add(i); for (int x : zm) if (x != -1) napraw(x); t1.update(i, m1[c - 'a']); cout << rozw() << '\n'; } } |
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