HDU5003 Osu! (Question G of the Anshan division Network Competition in 2014), hdu5003osu

HDU5003 Osu! (Question G of the Anshan division Network Competition in 2014), hdu5003osu

1. Question Description: Click to open the link

2. Problem-Solving ideas: This question is a simple sorting question, which can be calculated based on the meaning of the question.

3. Code:

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<functional>using namespace std;#define me(s) memset(s,0,sizeof(s))#define pb push_backtypedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;typedef pair <int, int> P;const int N=55;int a[N];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        sort(a,a+n);        double tot=0.0;        for(int i=0;i<n;i++)        {            int j=n-1-i;            tot+=(double)pow(0.95,i)*a[j];        }        printf("%.10lf\n",tot);    }}

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