题目链接
利用期望的线性性质:
(E(sum) = E(x_l) + E(x_{l+1})+ E(x_{l+2}) +.. E(x_r))
然后就考虑对于交换时两个区间元素的改动.
假设这两个区间的长度分别为(L_1,L_2),和为(S_1,S_2)
那么对于第一个区间元素的期望为.
(E(x_i) = dfrac{L_1 - 1}{L_1} * E(x_i) + dfrac{1}{L_1}*dfrac{L_2}{S_2})
对于第二个区间也是这样.
然后用线段树维护一下就可以了.
#include <iostream>
#include <cstdio>
#define lson now << 1
#define rson now << 1 | 1
const int maxN = 1e5 + 7;
struct Node {
int l,r;
double sum,lazy,mul;
}tree[maxN << 2];
inline int read() {
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
void init(int now,int l,int r) {
tree[now].l = l;tree[now].r = r;
tree[now].mul = 1;tree[now].lazy = 0;
return ;
}
void updata(int now) {
tree[now].sum = tree[lson].sum + tree[rson].sum;
return;
}
void build(int now,int l,int r) {
init(now,l,r);
if(l == r) {
tree[now].sum = read();
return;
}
int mid = (l + r) >> 1;
build(lson,l,mid);
build(rson,mid + 1,r);
updata(now);
return ;
}
void work(int now,double lazy,double mul) {
tree[now].lazy = tree[now].lazy * mul + lazy;
tree[now].mul = tree[now].mul * mul;
tree[now].sum = tree[now].sum * mul + lazy * (tree[now].r - tree[now].l + 1);
return ;
}
void pushdown(int now) {
work(lson,tree[now].lazy,tree[now].mul);
work(rson,tree[now].lazy,tree[now].mul);
tree[now].lazy = 0;tree[now].mul = 1;
return ;
}
double query(int now,int L,int R) {
if(tree[now].l >= L && tree[now].r <= R) return tree[now].sum;
int mid = (tree[now].l + tree[now].r) >> 1;
double sum = 0;
if(tree[now].lazy || tree[now].mul != 1) pushdown(now);
if(mid >= L) sum += query(lson,L,R);
if(mid < R) sum += query(rson,L,R);
return sum;
}
void modify(int now,int L,int R,double lazy,double mul) {
if(tree[now].l >= L && tree[now].r <= R) {
work(now,lazy,mul);
return ;
}
if(tree[now].lazy || tree[now].mul != 1) pushdown(now);
int mid = (tree[now].l + tree[now].r) >> 1;
if(mid >= L) modify(lson,L,R,lazy,mul);
if(mid < R) modify(rson,L,R,lazy,mul);
updata(now);
return ;
}
int main() {
int n,m;
n = read();m = read();
build(1,1,n);
int opt,l1,r1,l2,r2;
while(m --) {
opt = read() - 1;l1 = read();r1 = read();
if(!opt) {
l2 = read();r2 = read();
int L_1,L_2;
double S_1,S_2;
L_1 = r1 - l1 + 1;L_2 = r2 - l2 + 1;
S_1 = query(1,l1,r1);S_2 = query(1,l2,r2);
modify(1,l1,r1,1.0 / L_1 * S_2 / L_2,1.0 * (L_1 - 1) / L_1);
modify(1,l2,r2,1.0 / L_2 * S_1 / L_1,1.0 * (L_2 - 1) / L_2);
}
else printf("%.7lf
",query(1,l1,r1));
}
return 0;
}