XTOJ 1249 Rolling Variance【数学，滚动方差】_滚动方差什么意思-优快云博客

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/53011344

博客详细解析了XTOJ 1249题目的滚动方差问题，指出常规方法会超时，并解释了原因。通过公式推导，说明了求滚动方差时的效率问题，并提供了参考链接和AC代码。

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   Rolling Variance 
Accepted : 68 Submit : 197
Time Limit : 3000 MS Memory Limit : 65536 KB Special Judge
 

 
  Rolling VarianceBobo 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} 
         .
 InputThe 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 
         ).
 OutputFor 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 
       
 

原题链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1249

题意：如题目名字一样，求滚动方差，运用一般的方法会超时，就算预处理求出前n项和也还是会超时。

原因如下：

∑i=1n(ai−a¯)2=∑i=1na2i−∑i=1n2aia¯+∑i=1na¯2=∑i=1na2i−2∗a¯∗∑i=1nai+n∗a¯2

参考博客：http://blog.youkuaiyun.com/hnust_V/article/details/51763129

AC代码：

/**
  * 行有余力，则来刷题！
  * 博客链接：http://blog.youkuaiyun.com/hurmishine
  *
*/

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int maxn=1e5+5;
int a[maxn];
int sum[maxn];
int sum2[maxn];
int main()
{
    int n,m;
    //freopen("data.txt","r",stdin);
    while(cin>>n>>m)
    {
        memset(sum,0,sizeof(sum));
        memset(sum2,0,sizeof(sum2));
        for(int i=1; i<=n; i++)
        {
            cin>>a[i];
            sum[i]+=sum[i-1]+a[i];
            sum2[i]+=sum2[i-1]+a[i]*a[i];
        }
        /**
        两种方法均可
        for(int i=m; i<=n; i++)
        {
            double avg=1.0*(sum[i]-sum[i-m])/m;
            double ans=sqrt((sum2[i]-sum2[i-m]-2*avg*(sum[i]-sum[i-m])+m*avg*avg)/(m-1));
            printf("%.8f\n",ans);
        }
        */
        for(int i=0;i<=n-m;i++)
        {
            double avg=1.0*(sum[i+m]-sum[i])/m;
            double ans=sqrt((sum2[i+m]-sum2[i]-2*avg*(sum[i+m]-sum[i])+m*avg*avg)/(m-1));
            printf("%.8lf\n",ans);
        }
    }
    return 0;
}

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