【解題】Above Average

ACM Volume CIII 10370 - Above Average


The Problem

It is said that 90% of frosh expect to be above average in their class. You are to provide a reality check.


Input

The first line of standard input contains an integer C, the number of test cases. C data sets follow. Each data set begins with an integer, N, the number of people in the class (1 <= N <= 1000). N integers follow, separated by spaces or newlines, each giving the final grade (an integer between 0 and 100) of a student in the class.


Output

For each case you are to output a line giving the percentage of students whose grade is above average, rounded to 3 decimal places.


Sample Input

5
5 50 50 70 80 100
7 100 95 90 80 70 60 50
3 70 90 80
3 70 90 81
9 100 99 98 97 96 95 94 93 91


Sample Output

40.000%
57.143%
33.333%
66.667%
55.556%


解題思考

　　這一題只是給與一定數量的「分數」，要求你計算出比全體平均分數高的百分率而已。

　　首先，我們在輸入的分數的同時，計算出總分之後，就是要計算比平均高的人數。

　　為求精準，我是將每個人的分數乘以人數再與總分比較，而不是先由總分求出平均分數再跟每個人的分數比較。

　　最後，根據人數求出百分率。完成！


參考解答(C++)

#include <iostream>
#include <iomanip>

using namespace std;

int main(void)
{
    int c;
    cin >> c;

    // 設定顯示至小數點第三位
    cout << setiosflags(ios::fixed) << setprecision(3);

    for (int i = 0; i < c; i++)
    {
        int n, *score, total = 0;
        cin >> n;

        // 輸入分數, 並計算總和
        score = new int[n];
        for (int j = 0; j < n; j++)
        {
            cin >> score[j];

            total += score[j];
        }

        // 得到有多少人比平均高分
        int man = 0;
        for (int j = 0; j < n; j++)
        {
            if (score[j] * n > total)
            {
                man++;
            }
        }

        cout << (((double)man * 100) / n) << "%" << endl;

        delete [] score;
    }

#ifndef ONLINE_JUDGE
    system("pause");
#endif
}