萌新的比较工整的代码

解题思路:

注意事项:

（1）注意double精度不够，用long double

（2）避开所有人钱都一样的坑，直接输出0

参考代码:

#include <iostream> 

#include <vector>

#include <iomanip>

#include <cmath>

using namespace std;

typedef long double ld;

main()

{

        int n;cin>>n;

        ld s;cin>>s;

        vector<ld> a;

        ld x;

        for(int i=0;i<n;i++) cin>>x,a.push_back(x);

        bool flag=true;

        for(int i=1;i<n;i++)

                if(a[i]!=a[0]) {flag=false;break;}

        if(flag)

        {

            cout<<fixed<<setprecision(4)<<0.0;

            exit(0);

        }

        vector<ld> pay; 

        while(true) 

        {

                int len=a.size();

                ld avg=s/len;

                vector<ld> na;

                int count=0;

                for(int i=0;i<len;i++)

                if(a[i]>avg) na.push_back(a[i]);

                else s-=a[i],count++,pay.push_back(a[i]);

                a.clear();a=na;

                if(count==0) break;

        }

        int len=a.size();

        ld avg=s/len;

        for(int i=0;i<len;i++) pay.push_back(avg);

        ld sum=0,sumd=0;

        for(int i=0;i<n;i++) 

        sum+=pay[i],sumd+=pay[i]*pay[i];

        sum/=n;sumd/=n;

        ld r=abs(sumd-sum*sum);

        cout<<fixed<<setprecision(4)<<sqrt(r);

}