题解 | #字符统计#

s = input().strip()
s2 = set(s) # 去重
c = {}  # 存 i:count(i)的键值对
d = []  # 存所有的count(i)
for i in s2:
    if s.count(i) not in c:
        c[i] = s.count(i)
    else:
        c[i] = s.count(i) + 1
d = c.values()
e = sorted(set(d), reverse=True) # 所有的count(i)去重后倒序排列
output = ''
# 获取字典c中value = count(j)的所有keys
for j in e:
    sn = sorted([key for key, value in c.items() if value == j])
    output += ''.join(sn)
print(output)