CF808B

/*E - Average Sleep Time CodeForces - 808BIt's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days!When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last n days. So now he has a sequence a1, a2, ..., an, where ai is the sleep time on the i-th day.The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider k consecutive days as a week. So there will be n - k + 1 weeks to take into consideration. For example, if k = 2, n = 3 and a = [3, 4, 7], then the result is .You should write a program which will calculate average sleep times of Polycarp over all weeks.InputThe first line contains two integer numbers n and k (1 ≤ k ≤ n ≤ 2·105).The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).OutputOutput average sleeping time over all weeks.The answer is considered to be correct if its absolute or relative error does not exceed 10 - 6. In particular, it is enough to output real number with at least 6 digits after the decimal point.ExampleInput3 23 4 7Output9.0000000000Input1 110Output10.0000000000Input8 21 2 4 100000 123 456 789 1Output28964.2857142857NoteIn the third example there are n - k + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.尺取法*/#include<iostream>#include<queue>#include<cstring>#include<vector>#include<string>#include<algorithm>#include<cstdio>#include<cmath>#include<set>#include<cstdlib>#include<stack>#include<map>#include<functional>#define make_pair m_p#define deb cout<<"T"<<endlusing namespace std;typedef long long LL;const int maxsize = 1e6+7;const int INF = 0x3f3f3f3f;const int EXP = 1e-8;LL A[maxsize],B[maxsize];int n,k;int main(){    freopen("finput.txt","r",stdin);    while(~scanf("%d%d",&n,&k))    {        double sum = 0;        memset(A,0,sizeof A);memset(B,0,sizeof B);        for(int i=0;i<n;i++)        {            scanf("%lld",A+i);        }        for(int i=0;i<k;i++)        {            B[0]+=A[i];        }        sum += B[0];        for(int i=1;i<n-k+1;i++)        {            B[i] = B[i-1]+A[i+k-1] - A[i-1];            sum += B[i];        }        printf("%.10f\n",sum/(n-k+1));    }    return 0;}