Codeforces--808B

Average Sleep Time

题目链接

题意：大概就是求sum(ai+...ai+n-k)/(n-k+1).....(i=1--k)

题解：直接维护前缀和，带入公式即可，注意用long long保证精度，double会 损失精度

Code:

#include<bits/stdc++.h>#define ll long longll sum[2*100005],a[2*100005];int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        for(int i=1; i<=n; i++)        {            scanf("%I64d",&a[i]);            sum[i]=sum[i-1]+a[i];        }        ll res=0;        sum[0]=0;        for(int i=k; i<=n; i++)        {            res+=(sum[i]-sum[i-k]);        }        printf("%.10lf\n",(double)res/(n-k+1));    }    return 0;}