BAEKJOON PYTHON 2108 통계학

Prob

---

Approach

두 번째로 작은 값을 찾기 위해 index를 사용한다.

---

Code

import sys

N = int(sys.stdin.readline())

num = []
m_num = [0] * 8001
for i in range(N):
    num.append(int(sys.stdin.readline()))
    m_num[num[i] + 4000] += 1
num.sort()

print("%d" % (round(sum(num) / len(num))))
print("%d" % (num[len(num)//2]))

if m_num.count(max(m_num)) > 1:
    print(m_num.index(max(m_num), m_num.index(max(m_num)) + 1, len(m_num)) - 4000)
else:
    print(m_num.index(max(m_num)) - 4000)

print("%d" % (abs(max(num) - min(num))))

 

출처

https://www.acmicpc.net/problem/2108