C++14(g++5.4) 解法, 执行用时: 116ms, 内存消耗: 2144K, 提交时间: 2019-10-06 16:17:26

#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define N 1000100

double A[N],sum1,S;
int n,m;

int main () {
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>A[i];
	for(int i=1;i<=m;i++) sum1+=A[i]*A[i],S+=A[i];
	printf("%.8lf\n",sqrt((sum1-S*S/m)/(m-1)));
	for(int i=1;i<=n-m;i++) {
		sum1+=A[i+m]*A[i+m];
		sum1-=A[i]*A[i];
		S+=A[i+m];
		S-=A[i];
		printf("%.8lf\n",sqrt((sum1-S*S/m)/(m-1)));
	}
}

C++11(clang++ 3.9) 解法, 执行用时: 97ms, 内存消耗: 3560K, 提交时间: 2020-02-16 22:37:27

#include<cstdio>
#include<cmath>
using namespace std;
int n,m,i;
double a[100100],b[100100],c[100100];
double ave,ans;
int main()
{
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
	{
		scanf("%lf",&a[i]);
		b[i]=b[i-1]+a[i];
		c[i]=c[i-1]+a[i]*a[i];
	}
	for(i=m;i<=n;i++)
	{
		ave=(b[i]-b[i-m])/m;
		ans=(c[i]-c[i-m]+ave*ave*m-2*ave*(b[i]-b[i-m]))/(m-1);
		printf("%.8f\n",sqrt(ans));
	}
}