H — Variance

Time Limit：1s Memory Limit：128MByte

Submissions：259Solved：80

DESCRIPTION

An array with length n is given.
You should support 2 types of operations.

1 x y change the x-th element to y.
2 l r print the variance of the elements with indices l, l + 1, … , r.
As the result may not be an integer, you need print the result after multiplying (r-l+1)^2.

The index of the array is from 1.

INPUT The first line is two integer n, m(1 <= n, m <= 2^{16}), indicating the length of the array and the number of operations.The second line is n integers, indicating the array. The elements of the array

0<=ai<=1040<=ai<=104.The following m lines are the operations. It must be either1 x yor2 l r OUTPUT For each query, print the result. SAMPLE INPUT 4 41 2 3 42 1 41 2 42 1 42 2 4 SAMPLE OUTPUT 20242

题意：给定n个值，每次可以改变其中某个值，也可以询问其中一个区间的方差。输出每次询问的结果。

思路：线段树。需要用到方差公式Dx=E(X^2)-(E(X))^2。结点储存该区间内平方和及和。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <math.h>
using namespace std;
#define maxn 65540

struct node{
	int l,r;
	long long sum2,sum;
}n[4*maxn];

long long a[maxn];
long long sum2;
long long sum;

void build(int l,int r,int i){
	n[i].l=l;
	n[i].r=r;
	if(n[i].l==n[i].r){
		n[i].sum=a[l];
		n[i].sum2=a[l]*a[l];
		return;
	}
	int mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
	n[i].sum=n[2*i].sum+n[2*i+1].sum;
	n[i].sum2=n[2*i].sum2+n[2*i+1].sum2;
}

void update(int num,int s,int i){
	if(n[i].l==num&&n[i].r==num){
		n[i].sum=s;
		n[i].sum2=s*s;
		return;
	}
	int mid=(n[i].l+n[i].r)/2;
	if(num<=mid)
		update(num,s,2*i);
	else
		update(num,s,2*i+1);
	n[i].sum=n[2*i].sum+n[2*i+1].sum;
	n[i].sum2=n[2*i].sum2+n[2*i+1].sum2;
}

void query(int l,int r,int i){
	if(n[i].l==l&&n[i].r==r){
		sum2+=n[i].sum2;
		sum+=n[i].sum;
		return;
	}
	int mid=(n[i].l+n[i].r)/2;
	if(r<=mid)
		query(l,r,2*i);
	else if(l>mid)
		query(l,r,2*i+1);
	else{
		query(l,mid,2*i);
		query(mid+1,r,2*i+1);
	} 
}

int main(){
	
	int n,m;
	while(~scanf("%d %d",&n,&m)){
		for(int i=1;i<=n;i++){
			scanf("%lld",&a[i]);
		}
		build(1,n,1);
		while(m--){
			int q;
			scanf("%d",&q);
			if(q==1){
				int num,y;
				scanf("%d %d",&num,&y);
				update(num,y,1);
			}
			else{
				int l,r;
				sum2=0;
				sum=0;
				scanf("%d %d",&l,&r);
				query(l,r,1);
				long long ans=(r-l+1)*sum2-sum*sum;
				printf("%lld\n",ans);
			}
		}
	}
	return 0;
}