题目链接  Eyes Closed

题意  两个人玩一个游戏,现在有两种操作:

1、两个人格子挑选一个区间,保证两个的区间不相交。在这两个区间里面各选出一个数,交换这两个数。

2、挑选一个区间,求这个区间的和的期望。

对于第一种操作,先求出两个区间的长度$len1$和$len2$,再求出两个区间的期望和$s1$和$s2$。

对于第一个区间,我们先把这个区间里的所有数(期望值)乘上$(len1 - 1)/(len1)$,再加上$s2/len1/len2$

对于第二个区间,我们先把这个区间里的所有数(期望值)乘上$(len2 - 1)/(len2)$,再加上$s1/len1/len2$

线段树维护这两个操作即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
#define ls (i << 1)
#define rs (i << 1 | 1)
#define mid ((L + R) >> 1)
#define lson i << 1, L, mid
#define rson i << 1 | 1, mid + 1, R typedef long long LL; const int N = 4e5 + 10; double add[N], mul[N], s[N];
int n, q; void pushup(int i){ s[i] = s[ls] + s[rs]; } void pushdown(int i, int L, int R){
s[ls] = mul[i] * s[ls] + add[i] * (mid - L + 1);
mul[ls] *= mul[i];
add[ls] = mul[i] * add[ls] + add[i];
s[rs] = mul[i] * s[rs] + add[i] * (R - mid);
mul[rs] *= mul[i];
add[rs] = mul[i] * add[rs] + add[i];
add[i] = 0;
mul[i] = 1.0;
} void build(int i, int L, int R){
add[i] = 0, mul[i] = 1;
s[i] = 0;
if (L == R){ scanf("%lf", s + i); return; }
build(lson);
build(rson);
pushup(i);
} void add_update(int i, int L, int R, int l, int r, double val){
if (l <= L && R <= r){
s[i] += (R - L + 1) * val;
add[i] += val;
return;
} pushdown(i, L, R);
if (l <= mid) add_update(lson, l, r, val);
if (r > mid) add_update(rson, l, r, val);
pushup(i);
} void mul_update(int i, int L, int R, int l, int r, double val){
if (l <= L && R <= r){
s[i] *= val;
mul[i] *= val;
add[i] *= val;
return;
} pushdown(i, L, R);
if (l <= mid) mul_update(lson, l, r, val);
if (r > mid) mul_update(rson, l, r, val);
pushup(i);
} double query(int i, int L, int R, int l, int r){
if (l <= L && R <= r) return s[i];
double ret = 0;
pushdown(i, L, R);
if (l <= mid) ret += query(lson, l, r);
if (r > mid) ret += query(rson, l, r);
pushup(i);
return ret;
} int main(){ scanf("%d%d", &n, &q);
build(1, 1, n); while (q--){
int op;
scanf("%d", &op);
if (op == 1){
int l1, r1, l2, r2;
scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
double s1 = query(1, 1, n, l1, r1);
double s2 = query(1, 1, n, l2, r2);
double len1 = r1 - l1 + 1;
double len2 = r2 - l2 + 1;
mul_update(1, 1, n, l1, r1, 1.0 * (len1 - 1) / len1);
mul_update(1, 1, n, l2, r2, 1.0 * (len2 - 1) / len2);
add_update(1, 1, n, l1, r1, 1.0 / len1 / len2 * s2);
add_update(1, 1, n, l2, r2, 1.0 / len1 / len2 * s1);
} else{
int l, r;
scanf("%d%d", &l, &r);
printf("%.12f\n", query(1, 1, n, l, r));
}
} return 0;
}
05-28 16:35