Varying Kibibits
题意翻译
对于一个有$n$ 个元素的序列$a_1,a_2...a_n$ ，我们姑且把它称之为$T$
然后对于一个非空序列$L$ ，我们定义函数$f(L)$ :
将$L$ 中的元素十进制下的高位用0补全，取每一位的最小值组成一个新数，作为函数的值。
定义函数$G(x)$
$$G(x)=x\left(((\sum_{S \subseteq T,S \neq \emptyset,f(S)=x} (\sum_{y \in S}y)^2) \bmod 1000000007\right)$$
求$G(0) xor G(1) xor ... xor G(999999)$
感谢@litble 提供的翻译
题目描述
You are given $ n $ integers $ a_{1},a_{2},...,a_{n} $ . Denote this list of integers as $ T $ .
Let $ f(L) $ be a function that takes in a non-empty list of integers $ L $ .
The function will output another integer as follows:
- First, all integers in $ L $ are padded with leading zeros so they are all the same length as the maximum length number in $ L $ .
- We will construct a string where the $ i $ -th character is the minimum of the $ i $ -th character in padded input numbers.
- The output is the number representing the string interpreted in base 10.
For example $ f(10,9)=0 $ , $ f(123,321)=121 $ , $ f(530,932,81)=30 $ .
Define the function
 Here,  denotes a subsequence.In other words, $ G(x) $ is the sum of squares of sum of elements of nonempty subsequences of $ T $ that evaluate to $ x $ when plugged into $ f $ modulo $ 1000000007 $ , then multiplied by $ x $ . The last multiplication is not modded.
You would like to compute $ G(0),G(1),...,G(999999) $ . To reduce the output size, print the value , where  denotes the bitwise XOR operator.
输入输出格式
输入格式
The first line contains the integer $ n $ ( $ 1<=n<=1000000 $ ) — the size of list $ T $ .
The next line contains $ n $ space-separated integers, $ a_{1},a_{2},...,a_{n} $ ( $ 0<=a_{i}<=999999 $ ) — the elements of the list.
输出格式
Output a single integer, the answer to the problem.
输入输出样例
输入样例 #1
3
123 321 555
输出样例 #1
292711924
输入样例 #2
1
999999
输出样例 #2
997992010006992
输入样例 #3
10
1 1 1 1 1 1 1 1 1 1
输出样例 #3
28160
说明
For the first sample, the nonzero values of $ G $ are $ G(121)=144611577 $ , $ G(123)=58401999 $ , $ G(321)=279403857 $ , $ G(555)=170953875 $ . The bitwise XOR of these numbers is equal to $ 292711924 $ .
For example, , since the subsequences $ [123] $ and $ [123,555] $ evaluate to $ 123 $ when plugged into $ f $ .
For the second sample, we have 
For the last sample, we have , where  is the binomial coefficient.