CF #737（div2）C. Moamen and XOR，位运算，结论题，推公式

problem

C. Moamen and XOR
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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).

Input
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).

Output
For each test case, print a single value — the number of different arrays that Moamen wins with.

Print the result modulo 1000000007 (109+7).

Example
inputCopy
3
3 1
2 1
4 0
outputCopy
5
2
1
Note
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].

solution

题意：

• 给出n和k（<2e5），可以构造任意长度为n且满足ai<2^k的序列，求这些序列中满足全部相&大于等于全部异或（即a1&a2&a3…&an >= a1^a2^a3^…^an的个数），答案%(1e9+7)

思路：

• 看到2e5的范围，猜想是个结论题，所以打表找规律。
• 奇数很容易找到，偶数需要额外推一下公式

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
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;
}
LL inv(LL x, LL p){ return mpow(x,p-2);}

int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int T;  cin>>T;
	while(T--){
		int n, k;  cin>>n>>k;
		LL t1 = mpow(2,n-1), t2 = 2*t1%mod;
		if(n%2==1)cout<<mpow(t1+1,k)<<"\n";//奇数
		else cout<<((2*mpow(t2,k)+t2*mpow(t1-1,k))%mod*inv(t2+2,mod))%mod<<"\n";//偶数
	}
	return 0;
}




  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/119581546