Codeforces Round #447 (Div. 2) B. Ralph And His Magic Field(思维)

Description

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

題目大意

在n*m得格子里放置整数，要求每一行每一列的乘积都为k，求有多少种不同的方法（%1000000007）。

解题思路

我们可以先留出最后一行和最后一列，因为无论前面n-1行和m-1列放置的乘积状态如何，我们都可以在最后一行（最后一列）乘以1或-1使其结果为k。所以放置的种类数即为 2(n−1)∗(m−1) 。直接(n-1)*(m-1)会爆ll，要两次快速幂计算结果。
特判：当k=-1时，如果n和m一个为偶数，一个为奇数结果为0，因为此时无法实现每一行每一列-1的个数为奇数个。

代码实现

#include
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll quick_pow(ll a,ll b)
{
    ll ans=1;
    while(b>0)
    {
        if(b%2)
            ans=((ans%mod)*(a%mod))%mod;
        a=((a%mod)*(a%mod))%mod;
        b/=2;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    ll k;
    ll n,m;
    cin>>n>>m>>k;
    if(k==-1&&(n%2)!=(m%2))
        cout<<"0"<else
        cout<2,n-1),m-1)<return 0;
}

PS：可用降幂公式对指数先进行处理。
Ax mod p=Ax mod φ(p)+φ(p) mod p ，该题中 φ(p)=p−1

ll t=((n-1)%(mod-1))*((m-1)%(mod-1))%(mod-1)+(mod-1);
cout<2,t)<