标签：数学，快速幂
题目
Ralph has a magic field which is divided into n × m blocks. That is to say, there are n rows and m columns 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
给定长宽为N*M的矩阵，每个单位格子内可以填充1或者-1，给定k（1 or -1），一种符合要求的填充方法为每行每列的乘积都是k，问有多少种符合要求的填充方法
分析
显然，每个填充方法只和每行每列的最后一个单位有关，前面的n-1行m-1列都可以任意填充，所以方案转化为$ 2^{(n-1)*(m-1)} $
当然有种特殊情况，当n和m的奇偶性不一致时，无符合条件的方案，直接输出-1
code
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
using namespace std;
inline ll read()
{
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const ll p=1e9+7;
ll n,m,k;
inline ll qpow(ll x,ll y){
ll re=1;
x=x%p;
while(y){
if(y&1)re=(re*x)%p;
x=(x*x)%p;
y>>=1;
}
return re;
}
int main()
{
cin>>n>>m>>k;
if(n%2!=m%2&&k==-1){cout<<"0";return 0;}
cout<<qpow(qpow(2,n-1),m-1);
return 0;
}
