AC代码
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 1010;
int n;
int w[N];
int main()
{
cin >> n;
int sum = 0;
for (int i = 0; i < n; i ++ )
{
cin >> w[i];
sum += w[i];
}
double avg = (double)sum / n;//此题没啥好注意的,注意精度即可
double d = 0;
for (int i = 0; i < n; i ++ )
d += pow(w[i] - avg, 2) / n;
d = sqrt(d);
for (int i = 0; i < n; i ++ )
cout << (w[i] - avg) / d << endl;
return 0;
}
这道题比较好理解,没有太多难的概念,按照步骤一步一步的算就行,只需要记得精度控制好就行