억억단을 외우자 풀이

Author: saebyeok0306

Programmers/Level3/억억단을 외우자.py

@@ -0,0 +1,50 @@
+# 연습문제
+# https://school.programmers.co.kr/learn/courses/30/lessons/138475
+
+def find(counts, desc, start):
+    for d in desc:
+        nums = counts[d]
+        for i in nums:
+            if i >= start:
+                return i
+
+def solution(e, starts):
+    from collections import defaultdict
+    divisor = [0 for _ in range(e+1)]
+    counts = defaultdict(list)
+    divisor[1] = 1
+
+    # 파이썬으로 통과하고 싶으면, 아래의 최적화 기법으로 계산해야됨.
+    # for i in range(1, int(e**0.5)+1):
+    #     divisor[i**2] += 1
+
+    # 이부분은 이해가 안됨...
+    # for i in range(2,e+1):
+    #     for j in range(1,min(e//i+1,i)):
+    #         divisor[i*j] += 2
+
+    for i in range(2, e+1):
+        for j in range(1, e//i+1):
+            divisor[i*j] += 1
+    
+    for i in range(1, e+1):
+        counts[divisor[i]].append(i)
+    
+    desc = sorted(counts.keys(), reverse=True)
+    answer = []
+    for start in starts:
+        res = find(counts, desc, start)
+        answer.append(res)
+
+    return answer
+
+e = [8]
+starts = [[1,3,7]]
+result = [[6,6,8]]
+
+for q in [0]:
+    qid = solution(e[q], starts[q])
+    if qid == result[q]:
+        print(f'correct {qid}')
+    else:
+        print(f'incorrect {qid}')