Variance

1063 - Variance

Time Limit：1s Memory Limit：128MByte

Submissions：332Solved：101

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$0<=a_i<={10}^4$.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 4

1 2 3 4

2 1 4

1 2 4

2 1 4

2 2 4

SAMPLE OUTPUT

20

24

2

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<stdio.h>#include<iostream>using namespace std;long long int a[70000],b[70000],c[70000];int n;long long int lowbit(long long int i){    return i&-i;}void add(long long int i,long long int num){    while(i<=n)    {        b[i]+=num;        i+=lowbit(i);    }}void add1(long long int i,long long int num){    while(i<=n)    {        c[i]+=num;        i+=lowbit(i);    }}long long int sum1(long long int i){    long long int sum=0;    while(i>0)    {        sum+=b[i];        i-=lowbit(i);    }    return sum;}long long int sum2(long long int i){    long long int sum=0;    while(i>0)    {        sum+=c[i];        i-=lowbit(i);    }    return sum;}int main(){    long long int m;    while(scanf("%d%lld",&n,&m)!=-1)    {        for(long long int i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            add(i,a[i]);            add1(i,a[i]*a[i]);        }        long long int ans;        while(m--)        {            long long int w,e,r;            scanf("%lld%lld%lld",&w,&e,&r);            if(w==1)            {                long long int ha;                ha=r-a[e];                add(e,ha);                long long int haha=r*r-a[e]*a[e];                add1(e,haha);                a[e]=r;            }        else {if(e>1)            ans=-(sum1(r)-sum1(e-1))*(sum1(r)-sum1(e-1))+(r-e+1)*(sum2(r)-sum2(e-1));else ans=-sum1(r)*sum1(r)+r*(sum2(r));printf("%lld\n",ans);            }            }    }}