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https://lydsy.com/JudgeOnline/problem.php?id=5020
answer
This operation is obviously appear and disappear forced together forced to ah increase in code length.
So what is the equivalent of two sets of LCT on the line.
So consider how quickly find a bunch of stuff and score.
\ (SiN, exp \) , an addition between the functions do not seem beautiful nature, so we consider the Taylor expansion.
\ [E ^ v = \ sum_
{i = 0} ^ {\ infty} \ frac1 {!} \ Cdot v ^ i \] We \ (v = ax + b \ ) with incorporated, it is
\ [\ the begin {align *} e ^ v & = \ sum_ {i = 0} ^ {\ infty} \ frac1 {i!} \ cdot (ax + b) ^ i \\ & = \ sum_ {i = 0} ^ {\ infty} \ frac1 {i!} \ sum_ {j = 0} ^ i \ binom ij a ^ jb ^ {ij} x ^ j \\ & = \ sum_ {j = 0} ^ {\ infty} a ^ jx ^ j \ sum_ {i = j}
^ {\ infty} \ frac {\ binom ij b ^ {ij}} {i!} \ end {align *} \] so that we can obtain each \ (x ^ i \) in front of a factor.
\ (sin \) , then the same token, do not re-write it again. If a function \ (x ^ i \) coefficients do not have front count.
Expand probably a dozen is enough here to open the \ (16 \) key.
Such is the case time complexity \ (O (m (16 \ 16 + log ^ n-2)) \) .
#include<bits/stdc++.h>
#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}
typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;
template<typename I> inline void read(I &x) {
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}
const int N = 100000 + 7;
const int M = 16;
#define lc c[0]
#define rc c[1]
int n, m;
ll fac[M], C[M][M];
struct Node {
int c[2], fa, rev;
double v[M], sum[M];
inline Node() {}
inline void set(const int &f, const double &a, const double &b) {
memset(v, 0, sizeof(v));
if (f == 1) {
double aa = 1;
for (int i = 0; i < M; ++i) {
double bb = 1;
for (int j = i; j < M; ++j, bb *= b)
if ((j & 1) && (j >> 1) & 1) v[i] -= C[j][i] * bb / fac[j];
else if ((j & 1) && !((j >> 1) & 1)) v[i] += C[j][i] * bb / fac[j];
v[i] *= aa, aa *= a;
}
} else if (f == 2) {
double aa = 1;
for (int i = 0; i < M; ++i) {
double bb = 1;
for (int j = i; j < M; ++j, bb *= b) v[i] += C[j][i] * bb / fac[j];
v[i] *= aa, aa *= a;
}
} else if (f == 3) v[0] = b, v[1] = a;
}
} t[N];
int st[N];
inline bool idtfy(int o) { return t[t[o].fa].rc == o; }
inline bool isroot(int o) { return t[t[o].fa].lc != o && t[t[o].fa].rc != o; }
inline void connect(int fa, int o, int d) { t[fa].c[d] = o, t[o].fa = fa; }
inline void pushup(int o) {
assert(o);
assert(!t[o].rev);
for (int i = 0; i < M; ++i)
t[o].sum[i] = t[t[o].lc].sum[i] + t[t[o].rc].sum[i] + t[o].v[i];
}
inline void pushdown(int o) {
if (!t[o].rev) return;
if (t[o].lc) t[t[o].lc].rev ^= 1, std::swap(t[t[o].lc].lc, t[t[o].lc].rc);
if (t[o].rc) t[t[o].rc].rev ^= 1, std::swap(t[t[o].rc].lc, t[t[o].rc].rc);
t[o].rev = 0;
}
inline void rotate(int o) {
assert(!isroot(o));
int fa = t[o].fa, pa = t[fa].fa, d1 = idtfy(o), d2 = idtfy(fa), b = t[o].c[d1 ^ 1];
if (!isroot(fa)) t[pa].c[d2] = o; t[o].fa = pa;
connect(o, fa, d1 ^ 1), connect(fa, b, d1);
pushup(fa), pushup(o);
assert(!t[0].lc && !t[0].rc);
}
inline void splay(int o) {
int x = o, tp = 0;
st[++tp] = x;
while (!isroot(x)) st[++tp] = x = t[x].fa;
while (tp) pushdown(st[tp--]);
while (!isroot(o)) {
int fa = t[o].fa;
if (isroot(fa)) rotate(o);
else if (idtfy(o) == idtfy(fa)) rotate(fa), rotate(o);
else rotate(o), rotate(o);
}
}
inline void access(int o) {
for (int x = 0; o; o = t[x = o].fa)
splay(o), t[o].rc = x, pushup(o);
}
inline void mkrt(int o) {
access(o), splay(o);
t[o].rev ^= 1, std::swap(t[o].lc, t[o].rc);
}
inline int getrt(int o) {
access(o), splay(o);
while (pushdown(o), t[o].lc) o = t[o].lc;
return splay(o), o;
}
inline void link(int x, int y) {
mkrt(x);
if (getrt(y) != x) t[x].fa = y;
else assert(0);
}
inline void cut(int x, int y) {
mkrt(x), access(y), splay(y);
if (t[y].lc == x && !t[x].rc) t[x].fa = t[y].lc = 0, pushup(y);
else assert(0);
}
inline void work() {
while (m--) {
char opt[10];
scanf("%s", opt);
if (*opt == 'a') {
int x, y;
read(x), read(y);
++x, ++y;
link(x, y);
} else if (*opt == 'd') {
int x, y;
read(x), read(y);
++x, ++y;
cut(x, y);
} else if (*opt == 'm') {
double a, b;
int x, opt;
read(x), read(opt), scanf("%lf%lf", &a, &b);
++x;
splay(x), t[x].set(opt, a, b), pushup(x);
} else {
int x, y;
double v, vv = 1, ans = 0;
read(x), read(y), scanf("%lf", &v);
++x, ++y;
if (getrt(x) != getrt(y)) { puts("unreachable"); continue; }
mkrt(x), access(y), splay(y);
for (int i = 0; i < M; ++i, vv *= v) ans += vv * t[y].sum[i];
printf("%.8le\n", ans);
}
}
}
inline void init() {
read(n), read(m);
fac[0] = 1;
for (int i = 1; i < M; ++i) fac[i] = fac[i - 1] * i;
C[0][0] = 1;
for (int i = 1; i < M; ++i) {
C[i][0] = 1;
for (int j = 1; j < M; ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
int opt;
double a, b;
read(opt);
for (int i = 1; i <= n; ++i) read(opt), scanf("%lf%lf", &a, &b), t[i].set(opt, a, b), pushup(i);
}
int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif
init();
work();
fclose(stdin), fclose(stdout);
return 0;
}