1078 : Parity of Tuples (Easy)
时间限制Time Limit
5
秒Sec
内存限制Memory Limit
512
兆MB
提交次数Submitted
5 次Times
通过次数Solved
5 次Times
标准评测 Standard
从标准输入读入,结果写到标准输出。评测将输出拆成 token,与标准答案逐项比较,不按整段逐字节比对。
Read from standard input and write to standard output. The judge splits the output into tokens and compares them with the official answer; it does not compare raw bytes.
| 比对Compare | token 相同即通过。中间的空格、制表符、换行可多可少。Matching tokens pass. Extra spaces, tabs, or newlines between them are ignored. |
|---|---|
| 不同则错Differs | token 个数或内容不同即错误。23 与 2 3、02 与 2、2.0 与 2 均视为不同。A different token count or value is wrong. 23 vs 2 3, 02 vs 2, and 2.0 vs 2 all differ. |
| 输入Input | 标准输入,格式见题面。Standard input; format as in the statement. |
| 输出Output | 标准输出。Standard output. |
题目描述Description
Bobo has \(n\) \(m\)-tuple \(v_1, v_2, \dots, v_n\), where \(v_i = (a_{i, 1}, a_{i, 2}, \dots, a_{i, m})\). He wants to find \(\mathrm{count}(x)\) which is the number of \(v_i\) where \(a_{i, j} \wedge x\) has odd number of ones in its binary notation for all \(j\). Note that \(\wedge\) denotes the bitwise-and.
Find \(\sum_{x = 0}^{2^k - 1} \mathrm{count}(x) \cdot 3^x\) modulo \((10^9+7)\) for given \(k\).
输入格式Input
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains three integers \(n\), \(m\) and \(k\).
The \(i\)th of the following \(n\) lines contains \(m\) integers \(a_{i, 1}, a_{i, 2}, \dots, a_{i, m}\).
- \(1 \leq n \leq 10^4\)
- \(1 \leq m \leq 10\)
- \(1 \leq k \leq 30\)
- \(0 \leq a_{i, j} < 2^k\).
- There are at most \(100\) test cases, and at most \(1\) of them have \(n > 10^3\) or \(m > 5\).
输出格式Output
For each test case, print an integer which denotes the result.
样例Sample
出题Author
ftiasch