problem
B. AND 0, Sum Big
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Baby Badawy’s first words were “AND 0 SUM BIG”, so he decided to solve the following problem. Given two integers n and k, count the number of arrays of length n such that:
all its elements are integers between 0 and 2k−1 (inclusive);
the bitwise AND of all its elements is 0;
the sum of its elements is as large as possible.
Since the answer can be very large, print its remainder when divided by 109+7.
Input
The first line contains an integer t (1≤t≤10) — the number of test cases you need to solve.
Each test case consists of a line containing two integers n and k (1≤n≤105, 1≤k≤20).
Output
For each test case, print the number of arrays satisfying the conditions. Since the answer can be very large, print its remainder when divided by 109+7.
Example
inputCopy
2
2 2
100000 20
outputCopy
4
226732710
Note
In the first example, the 4 arrays are:
[3,0],
[0,3],
[1,2],
[2,1].
B. AND 0，总和
每次测试的时限2秒
每个测试的内存限制256 MB
输入标准输入
输出标准输出
Baby Badawy的第一个单词是“ AND 0 SUM BIG”，因此他决定解决以下问题。给定两个整数n和k，计算长度为n的数组的数量，使得：
它的所有元素都是0到2k-1（含）之间的整数；
其所有元素的按位与为0；
其元素的总和应尽可能大。
由于答案可能非常大，因此请打印除以109 + 7所得的余数。
输入
第一行包含一个整数t（1≤t≤10）-您需要解决的测试用例数。
每个测试用例由包含两个整数n和k（1≤n≤105，1≤k≤20）的一行组成。
输出
对于每个测试用例，打印满足条件的数组数。由于答案可能非常大，因此请打印除以109 + 7所得的余数。
例子
inputCopy
2
2 2
100000 20
outputCopy
4
226732710
笔记
在第一个示例中，这四个数组是：
[3,0]，
[0,3]，
[1,2],
[2,1]。
solution
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
const int mod = 1e9+7;
LL mpow(LL a, LL x) {
if(x==0)return 1;
LL t = mpow(a, x>>1);
if(x%2==0)return t*t%mod;
return t*t%mod*a%mod;
}
int main(){
ios::sync_with_stdio(false);
//init();
int T; cin>>T;
while(T--){
int n, k; cin>>n>>k;
cout<<mpow(n,k)<<"\n";
}
return 0;
}