Codeforces 895.E Eyes Closed

E. Eyes Closed

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya and Petya were tired of studying so they decided to play a game. Before the game begins Vasya looks at array a consisting of n integers. As soon as he remembers all elements of a the game begins. Vasya closes his eyes and Petya does q actions of one of two types:

1) Petya says 4 integers l1, r1, l2, r2 — boundaries of two non-intersecting segments. After that he swaps one random element from the [l1, r1]segment with another random element from the [l2, r2] segment.

2) Petya asks Vasya the sum of the elements of a in the [l, r] segment.

Vasya is a mathematician so he answers Petya the mathematical expectation of the sum of the elements in the segment.

Your task is to write a program which will answer the second type questions as Vasya would do it. In other words your program should print the mathematical expectation of the sum of the elements of a in the [l, r] segment for every second type query.

Input

The first line contains two integers n, q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105)  — the number of elements in the array and the number of queries you need to handle.

The second line contains n integers ai (1 ≤ ai ≤ 109)  — elements of the array.

The next q lines contain Petya's actions of type 1 or 2.

If it is a type 1 action then the line contains 5 integers 1, l1, r1, l2, r2 (1 ≤ l1 ≤ r1 ≤ n, 1 ≤ l2 ≤ r2 ≤ n).

If it is a type 2 query then the line contains 3 integers 2, l, r (1 ≤ l ≤ r ≤ n).

It is guaranteed that there is at least one type 2 query and segments [l1, r1], [l2, r2] don't have common elements.

Output

For each type 2 query print one real number — the mathematical expectation of the sum of elements in the segment.

Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4  — formally, the answer is correct if  where x is jury's answer and y is yours.

Examples

input

4 4
1 1 2 2
1 2 2 3 3
2 1 2
1 1 2 3 4
2 1 2

output

3.0000000
3.0000000

input

10 5
1 1 1 1 1 2 2 2 2 2
1 1 5 6 10
2 1 5
1 1 5 6 10
1 1 5 6 10
2 6 10

output

6.0000000
8.0400000

input

10 10
1 2 3 4 5 6 7 8 9 10
1 1 5 6 10
1 1 5 6 10
2 1 5
1 1 3 6 9
2 1 3
1 5 7 8 10
1 1 1 10 10
2 1 5
2 7 10
2 1 10

output

23.0000000
14.0000000
28.0133333
21.5733333
55.0000000

大致题意：两种操作。 
1. [l1, r1]之间随机一个数，[l2, r2]之间随机一个数，把两个交换 
2. 问[l, r]区间和的数学期望是多少。

分析：直接分析每个数对整体的影响很难，先分析个体.设len1 = r1 - l1 + 1,len2 = r2 - l2 + 1.那么对于第一个区间的数x，有1/len1的概率随机到.还有(len1 - 1)/len1的概率不会随机到.右边有1/len2的概率随机到y,那么既随机到x又随机到y的概率为1/(len1*len2),枚举第二个区间的每一个数y,那么x对第一个区间的期望的贡献就变成了[(len1 - 1)/len1] * x + sum2 / (len1 * len2).整个区间的期望和就是把所有x的期望加起来.可以利用线段树来维护：区间乘，区间加，区间求和，对于第二个区间也是差不多这样的.

对期望的概念理解的还不是非常深入.直接算整体的不好算可以先考虑算局部的贡献.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = ;

int n,q,L[maxn << ],R[maxn << ];
double sum[maxn << ],add[maxn << ],mul[maxn << ];

void pushup(int o)
{
    sum[o] = sum[o * ] + sum[o *  + ];
}

void pushdown(int o)
{
    sum[o * ] = sum[o * ] * mul[o] + add[o] * (R[o * ] - L[o * ] + );
    sum[o *  + ] = sum[o *  + ] * mul[o] + add[o] * (R[o *  + ] - L[o *  + ] + );
    add[o * ] = add[o * ] * mul[o] + add[o];
    add[o *  + ] = add[o *  + ] * mul[o] + add[o];
    mul[o * ] = mul[o * ] * mul[o];
    mul[o *  + ] = mul[o *  + ] * mul[o];
    mul[o] = ;
    add[o] = ;
}

void build(int o,int l,int r)
{
    L[o] = l;
    R[o] = r;
    add[o] = ;
    mul[o] = ;
    if (l == r)
    {
        cin >> sum[o];
        return;
    }
    int mid = (l + r) >> ;
    build(o * ,l,mid);
    build(o *  + ,mid + ,r);
    pushup(o);
}

double query(int o,int l,int r,int x,int y)
{
    if (x <= l && r <= y)
        return sum[o];
    pushdown(o);
    int mid = (l + r) >> ;
    double res = 0.0;
    if (x <= mid)
        res += query(o * ,l,mid,x,y);
    if (y > mid)
        res += query(o *  + ,mid + ,r,x,y);
    return res;
}

void update1(int o,int l,int r,int x,int y,double v)
{
    if (x <= l && r <= y)
    {
        sum[o] += v * (r - l + );
        add[o] += v;
        return;
    }
    pushdown(o);
    int mid = (l + r) >> ;
    if (x <= mid)
        update1(o * ,l,mid,x,y,v);
    if (y > mid)
        update1(o *  + ,mid + ,r,x,y,v);
    pushup(o);
}

void update2(int o,int l,int r,int x,int y,double v)
{
    if (x <= l && r <= y)
    {
        add[o] *= v;
        sum[o] *= v;
        mul[o] *= v;
        return;
    }
    pushdown(o);
    int mid = (l + r) >> ;
    if (x <= mid)
        update2(o * ,l,mid,x,y,v);
    if (y > mid)
        update2(o *  + ,mid + ,r,x,y,v);
    pushup(o);
}

int main()
{
    scanf("%d%d",&n,&q);
    build(,,n);
    while (q--)
    {
        int l1,r1,l2,r2;
        int id;
        scanf("%d",&id);
        if (id == )
        {
            scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
            double sum1 = query(,,n,l1,r1),sum2 = query(,,n,l2,r2);
            double len1 = r1 - l1 + ,len2 = r2 - l2 + ;
            update2(,,n,l1,r1,(len1 - ) / len1);
            update2(,,n,l2,r2,(len2 - ) / len2);
            update1(,,n,l1,r1,1.0 / len1 * (sum2 / len2));
            update1(,,n,l2,r2,1.0 / len2 * (sum1 / len1));
        }
        else
        {
            scanf("%d%d",&l1,&r1);
            printf("%.7lf\n",query(,,n,l1,r1));
        }
    }

    return ;
}