codeforces 894B. Ralph And His Magic Field
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题意:给你一个N*M的矩阵,矩阵中只放1和-1,问有多少种方案可以使每一行、每一列的乘积都为K(K为±1)
思路:不管(N - 1) * (M - 1)的矩阵中怎么填,我们都可以通过最后一行和最后一列决定这一行或这一列的值为1还是-1,所以答案共有种
注意当N,M一奇一偶且K为-1时,答案为0。因为不管怎么放,最后一行与最后一列一个有奇数个-1和一个有偶数个-1总会冲突。
N,M都是10^18,N*M会long long,所以应该进行两次快速幂。
#include<bits/stdc++.h>using namespace std;const long long MOD = 1e9 + 7;long long q_pow(long long a, long long b){long long res = 1;while(b){if(b&1) res = (res * a) % MOD;a = (a * a) % MOD;b = b >> 1;}return res % MOD;}int main(){long long n, m, k;while(~scanf("%lld%lld%lld", &n, &m, &k)){if((n % 2 + m % 2) % 2 && k == -1) printf("0\n");else printf("%lld\n",q_pow(q_pow(2, n - 1), m - 1));} return 0;}
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