Luogu4546 THUWC2017 在美妙的数学王国中畅游 LCT泰勒展开

Posted 2021-01-23 itst

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题意：反正就是一堆操作

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LCT总是和玄学东西放在一起
我们不妨令$x_0=0.5$（其实取什么都是一样的，但是最好取在$[0,1]$的范围内），将其代入给出的式子，我们得到的$f(x)$的式子就是一个多项式了。
然后复习一下导数：
$(Cf(x))‘=Cf‘(x)$（$C$为常数）
$sin‘(x)=cos(x),cos‘(x)=-sin(x),(e^x)‘=e^x,C‘=0 , (ax+b)‘=a$
令$g(x)=u$，则$f[g(x)]‘ = f‘(u) imes g‘(x)$
有了这些式子就可以得到给出的三种函数的任意阶导数了。
但是显然我们不能把所有项的系数都算出来。因为在比较靠后的项中，阶乘的值很大，对答案造成的贡献就会小到忽略不计，所以我们可以取前面若干项，这里我取的是前$12$项。
然后用$LCT$维护这些项的系数和，每一次询问把链拿出来直接算就行了，难题变成裸题了qwq

  1 #include<bits/stdc++.h>
  2 #define ld long double
  3 //This code is written by Itst
  4 using namespace std;
  5 
  6 inline int read(){
  7     int a = 0;
  8     bool f = 0;
  9     char c = getchar();
 10     while(c != EOF && !isdigit(c)){
 11         if(c == ‘-‘)
 12             f = 1;
 13         c = getchar();
 14     }
 15     while(c != EOF && isdigit(c)){
 16         a = (a << 3) + (a << 1) + (c ^ ‘0‘);
 17         c = getchar();
 18     }
 19     return f ? -a : a;
 20 }
 21 
 22 const int MAXN = 100010;
 23 struct node{
 24     ld point[13] , pre[13] , a , b;
 25     int ch[2] , fa , type;
 26     bool mark;
 27 }Tree[MAXN];
 28 int N;
 29 char s[20];
 30 
 31 inline bool nroot(int x){
 32     return Tree[Tree[x].fa].ch[1] == x || Tree[Tree[x].fa].ch[0] == x;
 33 }
 34 
 35 inline bool son(int x){
 36     return Tree[Tree[x].fa].ch[1] == x;
 37 }
 38 
 39 inline ld calc(int type , int n , ld a , ld b){
 40     ld sum = 0;
 41     switch(type){
 42     case 1:
 43         switch(n & 3){
 44         case 0:
 45             sum = sin(0.5 * a + b);
 46             break;
 47         case 1:
 48             sum = cos(0.5 * a + b);
 49             break;
 50         case 2:
 51             sum = -sin(0.5 * a + b);
 52             break;
 53         case 3:
 54             sum = -cos(0.5 * a + b);
 55             break;
 56         }
 57         return sum * pow(a , n);
 58     case 2:
 59         return pow(a , n) * exp(a * 0.5 + b);
 60     case 3:
 61         switch(n){
 62         case 0:
 63             return a * 0.5 + b;
 64         case 1:
 65             return a;
 66         default:
 67             return 0;
 68         }
 69     }
 70 }
 71 
 72 inline void pushup(int x){
 73     for(int i = 0 ; i <= 12 ; ++i)
 74         Tree[x].point[i] = Tree[x].pre[i] + Tree[Tree[x].ch[0]].point[i] + Tree[Tree[x].ch[1]].point[i];
 75 }
 76 
 77 inline void getpre(int x){
 78     for(int i = 0 ; i <= 12 ; ++i)
 79         Tree[x].pre[i] = calc(Tree[x].type , i , Tree[x].a , Tree[x].b);
 80 }
 81 
 82 inline void ZigZag(int x){
 83     bool f = son(x);
 84     int y = Tree[x].fa , z = Tree[y].fa , w = Tree[x].ch[f ^ 1];
 85     if(nroot(y))
 86         Tree[z].ch[son(y)] = x;
 87     Tree[x].fa = z;
 88     Tree[x].ch[f ^ 1] = y;
 89     Tree[y].fa = x;
 90     Tree[y].ch[f] = w;
 91     if(w)
 92         Tree[w].fa = y;
 93     pushup(y);
 94     pushup(x);
 95 }
 96 
 97 inline void pushdown(int x){
 98     if(Tree[x].mark){
 99         Tree[Tree[x].ch[0]].mark ^= 1;
100         Tree[Tree[x].ch[1]].mark ^= 1;
101         Tree[x].mark = 0;
102         swap(Tree[x].ch[0] , Tree[x].ch[1]);
103     }
104 }
105 
106 void pushdown_all(int x){
107     if(nroot(x))
108         pushdown_all(Tree[x].fa);
109     pushdown(x);
110 }
111 
112 inline void Splay(int x){
113     pushdown_all(x);
114     while(nroot(x)){
115         if(nroot(Tree[x].fa))
116             ZigZag(son(x) == son(Tree[x].fa) ? Tree[x].fa : x);
117         ZigZag(x);
118     }
119 }
120 
121 inline void access(int x){
122     for(int y = 0 ; x ; y = x , x = Tree[x].fa){
123         Splay(x);
124         Tree[x].ch[1] = y;
125         pushup(x);
126     }
127 }
128 
129 inline int findroot(int x){
130     access(x);
131     Splay(x);
132     pushdown(x);
133     while(Tree[x].ch[0])
134         pushdown(x = Tree[x].ch[0]);
135     Splay(x);
136     return x;
137 }
138 
139 inline void makeroot(int x){
140     access(x);
141     Splay(x);
142     Tree[x].mark ^= 1;
143 }
144 
145 inline void split(int x , int y){
146     makeroot(x);
147     access(y);
148     Splay(y);
149 }
150 
151 inline void link(int x , int y){
152     makeroot(x);
153     Tree[x].fa = y;
154 }
155 
156 inline void cut(int x , int y){
157     split(x , y);
158     Tree[y].ch[0] = Tree[x].fa = 0;
159     pushup(y);
160 }
161 
162 inline void change(int x , int type , ld a , ld b){
163     access(x);
164     Splay(x);
165     Tree[x].type = type;
166     Tree[x].a = a;
167     Tree[x].b = b;
168     getpre(x);
169     pushup(x);
170 }
171 
172 int main(){
173     freopen("4546.in" , "r" , stdin);
174     freopen("4546.out" , "w" , stdout);
175     N = read();
176     int M = read();
177     read();
178     for(int i = 1 ; i <= N ; ++i){
179         Tree[i].type = read();
180         scanf("%Lf %Lf" , &Tree[i].a , &Tree[i].b);
181         getpre(i);
182     }
183     while(M--){
184         ld a , b , times , jc;
185         int d , e;
186         if(scanf("%s" , s) == EOF)
187             return 0;
188         switch(s[0]){
189         case ‘a‘:
190             d = read() + 1;
191             e = read() + 1;
192             link(d , e);
193             break;
194         case ‘d‘:
195             d = read() + 1;
196             e = read() + 1;
197             cut(d , e);
198             break;
199         case ‘m‘:
200             d = read() + 1;
201             e = read();
202             scanf("%Lf %Lf" , &a , &b);
203             change(d , e , a , b);
204             break;
205         case ‘t‘:
206             d = read() + 1;
207             e = read() + 1;
208             scanf("%Lf" , &a);
209             if(findroot(d) != findroot(e))
210                 puts("unreachable");
211             else{
212                 split(d , e);
213                 b = 0;
214                 times = jc = 1;
215                 a -= 0.5;
216                 for(int i = 0 ; i <= 12 ; ++i){
217                     b += times * Tree[e].point[i] / jc;
218                     times *= a;
219                     jc *= (i + 1);
220                 }
221                 printf("%.9Lf
" , b);
222             }
223         }
224     }
225     return 0;
226 }