894B. Ralph And His Magic Field

那天晚上开题做的时候读懂了题意，但是没想到方法，今天晚上看别人博客，还有实验室的同学说，知道了做的方法了。然后利用学长给我讲的快速幂模板套上就可以了。当然网上的博主们的快速幂，代码更简洁，但是我已经理解了学长给我的快速幂，就懒得在学了….记性也不好….
这里有详细的过程http://blog.csdn.net/riba2534/article/details/78583483
B. Ralph And His Magic Field
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

    #include
    #include
    #include
    #include
    using namespace std;
    const long long mod=1e9+7;
    long long n,m;
    long long qpow(long long n,long long m)
    {
        if(m==0)
        return 1;
        if(m==1)
        return n;
        long long num=1;
        num=(qpow(n,m/2)*qpow(n,m/2))%mod;
        if(m&1)
        num=(num*n)%mod;
        return num;
}
    int main()
    {
    int k;
    while(cin>>n>>m>>k)
    {
    if(m%2!=n%2&&k==-1)
        cout<<"0"<

}