CodeForces - 894B Ralph And His Magic Field（思维+快速幂）

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.

Example
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：填数字，使每一行每一列的乘积都相等都等于k，k的值不是1就是-1

解题思路：k的值不是1就是-1所以填的数肯定也是不是1就是-1啦。
主要在于1和-1有奇数个还是偶数个。

我们可以把每一行的最后一列和每一列的最后一行空出来，用来凑成结果K.
举个例子：
如果k=-1
如果前n-1个数的乘积是1那最后一个格子填上-1就能使等式成立。

所以(n-1)*(m-1)个格子一共有2^(n-1)(m-1)种情况。

但并不是所有情况都能输出结果。
考虑当行和列的奇偶性不相等而k=-1时得不出结果。这就是在于-1有奇数个还是偶数个了，如果行和列的奇偶性不一致，那么如果能凑成行的话就一定凑不成列，同样，如果能凑成列的话就一定凑不成行。

最后注意数据规模，就要用到快速幂了。

AC代码：

#include
typedef long long ll;

ll mod_pow(ll x,ll n,ll mod)
{
    if(n==0) return 1;
    ll res=mod_pow(x*x%mod,n/2,mod);
    if(n&1) res=res*x%mod;
    return res;
}

int main()
{
    ll n,m;
    int k;
    while(~scanf("%lld%lld%d",&n,&m,&k))
    {
        if((n%2!=m%2)&&k==-1) {printf("0\n");continue;}
        printf("%lld\n",mod_pow(mod_pow(2,n-1,1000000007),m-1,1000000007));
    }
    return 0;
}