Finding the percentage 少数第二位まで出力する方法

・問題

You have a record of  students. Each record contains the student's name, and their percent marks in Maths, Physics and Chemistry. The marks can be floating values. The user enters some integer  followed by the names and marks for  students. You are required to save the record in a dictionary data type. The user then enters a student's name. Output the average percentage marks obtained by that student, correct to two decimal places.

Input Format

The first line contains the integer , the number of students. The next  lines contains the name and marks obtained by that student separated by a space. The final line contains the name of a particular student previously listed.

Constraints

 

Output Format

Print one line: The average of the marks obtained by the particular student correct to 2 decimal places.

Sample Input 0

3
Krishna 67 68 69
Arjun 70 98 63
Malika 52 56 60
Malika

Sample Output 0

56.00

Explanation 0

Marks for Malika are  whose average is 

Sample Input 1

2
Harsh 25 26.5 28
Anurag 26 28 30
Harsh

Sample Output 1

26.50

 

 

 

・ソースコード

if __name__ == '__main__':
    n = int(input())
    student_marks = {}
    for _ in range(n):
        name, *line = input().split()
        scores = list(map(float, line))
        student_marks[name] = scores
    query_name = input()
print ('%0.2f' %((((student_marks[query_name][0])+(student_marks[query_name][1])+(student_marks[query_name][2]))/3)))

・コメント

いつも通り、名前とスコアが入力として与えられる問題。

入力の最後にある名前の人の点数を平均して出力するだけなので難しくはないが、出力は少数第二位までなければならないのでそこが唯一難しいところ。

print ('%0.2f' %(なんちゃら))

という形にすれば出力できることさえ分かればできる。

・URL

https://www.hackerrank.com/challenges/finding-the-percentage/problem