import sys
# 对于个位数,每个5结尾的一定符合要求
# 对于其他数,首先其要是3的倍数,其次要是奇数
# 30一次循环,一轮循环7个 —— 3, 5, 9, 15, 21, 25, 27
class Solution(object):
    def get_kth_number(self, k: int):
        # 基本情况,小于7,即在一轮循环内
        basic = [3, 5, 9, 15, 21, 25, 27]
        if k <= 7:
            print(basic[k-1])
            return
            
        times = (k-1) // 7  # 循环轮数
        pos = (k-1) % 7
        print(30*times + basic[pos])

# 测试
if __name__ == "__main__":
    group = []
    for line in sys.stdin:
        group.append(line.strip())
    
    solut = Solution()
    for i in range(int(group[0])):
        solut.get_kth_number(int(group[i+1]))






数学问题,发现其7个一组,每组间隔30就很简单了