XTU 1249 Rolling Variance

Rolling Variance

Accepted : 77		Submit : 212
Time Limit : 3000 MS		Memory Limit : 65536 KB Special Judge

Rolling Variance

Bobo learnt that the variance of a sequence a1,a2,…,an is

∑ni=1(ai−a¯)2n−1−−−−−−−−−−−−√

where

a¯=∑ni=1ain.

Bobo has a sequence a1,a2,…,an,and he would like to find the variance of each consecutive subsequences of length m.Formally, the i-th (1≤i≤n−m+1) rolling variance ri is the variance of sequence {ai,ai+1,…,ai+m−1}.

Input

The input contains at most 30 sets. For each set:

The first line contains 2 integers n,m (2≤m≤n≤105).

The second line contains n integers a1,a2,…,an (|ai|≤100).

Output

For each set, (n−m+1) lines with floating numbers r1,r2,…,rn−m+1.

Your answer will be considered correct if its absolute or relative error does not exceed 10−4.

Sample Input

3 2
1 3 2
5 3
1 3 2 4 5

Sample Output

1.41421356
0.70710678
1.00000000
1.00000000
1.52752523

解题思路：先求出m个连续的数的平均值a，然后求(ai-a)^2

求和：(a0+a)^2+(a1+a)^2+(a2+a)^2+(a3+a)^2+···

           =a0^2+2*a*a0+a^2+a1^2+2*a*a1+a^2+a2^2+2*a*a2+a^2+·····

           =（a0^2+a1^2+····）+a2^2+2*a*（a0+a1+a2+····）+（a^2+a^2+a^2+·····）

代码如下：

#include 
#include 
#include 
#define maxn 100001
int a[maxn],b[maxn];
int main()
{
    int n,m,c;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        int i;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&c);
            a[i]=a[i-1]+c;///前i个数的和
            b[i]=b[i-1]+c*c;///前i个数的平方和
        }
        for(i=m; i<=n;i++)
        {
            double s=0;
            s=1.0*(a[i]-a[i-m])/(1.0*m);///求出平均数
            s=b[i]-b[i-m]+s*s*m-2.0*s*(a[i]-a[i-m]);
            printf("%0.8lf\n",(double)sqrt(s/(m-1)));
        }
    }
    return 0;
}

aa¯