CF1557C.Moamen and XOR
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
Moamen and Ezzat are playing a game. They create an array a of n non-negative integers where every element is less than 2k .
Moamen wins if a1&a2&a3&…&an≥a1⊕a2⊕a3⊕…⊕an .
Here & denotes the bitwise AND operation, and ⊕ denotes the bitwise XOR operation.
Please calculate the number of winning for Moamen arrays a .
As the result may be very large, print the value modulo 1000000007 ( 109+7 ).
输入格式
The first line contains a single integer t ( 1≤t≤5 )— the number of test cases.
Each test case consists of one line containing two integers n and k ( 1≤n≤2⋅105 , 0≤k≤2⋅105 ).
输出格式
For each test case, print a single value — the number of different arrays that Moamen wins with.
Print the result modulo 1000000007 ( 109+7 ).
输入输出样例
输入#1
3 3 1 2 1 4 0
输出#1
5 2 1
说明/提示
In the first example, n=3 , k=1 . As a result, all the possible arrays are [0,0,0] , [0,0,1] , [0,1,0] , [1,0,0] , [1,1,0] , [0,1,1] , [1,0,1] , and [1,1,1] .
Moamen wins in only 5 of them: [0,0,0] , [1,1,0] , [0,1,1] , [1,0,1] , and [1,1,1] .