F.A.Q
Hand In Hand
Online Acmers
Problem Archive
Realtime Judge Status
Authors Ranklist
 
     C/C++/Java Exams     
ACM Steps
Go to Job
Contest LiveCast
ICPC@China
Best Coder beta
VIP | STD Contests
    DIY | Web-DIY beta
Author ID 
Password 
 Register new ID

Find minimal sum

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 940    Accepted Submission(s): 639


Problem Description
每一个正整数都可以表示成m = An-1(n-1)! + An-2(n-2)! + An-3(n-3)! + ... + A1, (0<=Ai<=i, j =1, 2, ..., n-1)的形式. 但是对于一个给定的正整数, ACboy的老师要他找出最小的min(An-1 + An-2 + .. + A1)满足m = An-1(n-1)! + An-2(n-2)! + An-3(n-3)! + ... + A1.
你能帮帮他吗?
 

Input
输入首先给出一个N, 代表有N个测试实例。
接下来的N行, 每行包括一个正整数M (1 <= M < 2^32).

 

Output
对于每个测试实例输出min(An-1 + An-2 + .. + A1).
 

Sample Input
3 2 3 100
 

Sample Output
1 2 6
 

Hint
对于2, 可以写成3 = 1*2!, 所以答案为1.
对于3, 可以写成3 = 1*2! + 1*1!, 所以答案为2.
对于100, 可以写成100 = 4*4! + 2*2!, 所以答案为6.
 

Author
8600
 

Source
 

Statistic | Submit | Discuss | Note
Hangzhou Dianzi University Online Judge 3.0
Copyright © 2005-2024 HDU ACM Team. All Rights Reserved.
Designer & Developer : Wang Rongtao LinLe GaoJie GanLu
Total 0.000000(s) query 1, Server time : 2024-04-23 22:15:59, Gzip enabled