codeforces B. Ralph And His Magic Field 数学题+快速幂

最新推荐文章于 2020-09-10 20:36:19 发布

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最新推荐文章于 2020-09-10 20:36:19 发布

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探讨如何计算特定条件下n*m矩阵的填数方案总数，每行每列乘积为指定值，仅包含1和-1。利用矩阵特性及快速幂算法解决无限填数范围的问题。

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Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and mcolumns on the field. Ralph can put an integer in each block. However, the magic field doesn't always work properly. It works only if the product of integers in each row and each column equals to k, where k is either 1 or -1.

Now Ralph wants you to figure out the number of ways to put numbers in each block in such a way that the magic field works properly. Two ways are considered different if and only if there exists at least one block where the numbers in the first way and in the second way are different. You are asked to output the answer modulo 1000000007 = 109 + 7.

Note that there is no range of the numbers to put in the blocks, but we can prove that the answer is not infinity.

Input

The only line contains three integers n, m and k (1 ≤ n, m ≤ 1018, k is either 1 or -1).

Output

Print a single number denoting the answer modulo 1000000007.

Examples

input
1 1 -1

output
1

input
1 3 1

output
1

input
3 3 -1

output
16

Note

In the first example the only way is to put -1 into the only block.

In the second example the only way is to put 1 into every block.

题意：

给你n,m,k(k==-1 || k==1)，求出有多少个n*m的矩阵每行每列的乘积都为k。矩阵只包含1和-1。

思路：

根据矩阵的特点，随意给出一个(n-1)*(m-1)的矩阵，都可以配出一个n*m的目标矩阵。那么只需要求出(n-1)*(m-1)的矩阵有多少种就可以了。每个位置有2种取值，有(n-1)*(m-1)个位置，那么就有pow(2,(n-1)*(m-1))种矩阵。需要使用快速幂。当n,m一奇一偶时，不可能配成-1。

#include<cstdio>
#include<cstring>
#include<iostream>
#define maxn 100005
#define LL long long
#define mod 1000000007
using namespace std;
LL pow(LL u,LL n)
{
	LL x;
	x=1;
	while(n)
	{
		if(n&1)
		{
			x=x*u;
			x=x%mod;
		}
		u=u*u;
		u%=mod;
		n=n>>1;
	}
	return x;
}
int main()
{
	ios::sync_with_stdio(false);
	LL n,m,k;
	while(cin>>n>>m>>k)
	{
		if((n+m)%2==1&&k==-1)
		{
			cout << 0 << endl;
			continue;
		}
		LL ans;
		ans=pow(pow(2,n-1),m-1);
		ans%=mod;
		cout << ans << endl;
	}	
	return 0;	
}