Imagine A is a NxM matrix with two basic properties
1) Each element in the matrix is distinct and lies in the range of 1<=A[i][j]<=(N*M)
2) For any two cells of the matrix, (i1,j1) and (i2,j2), if (i1^j1) > (i2^j2) then A[i1][j1] > A[i2][j2] ,where
1 ≤ i1,i2 ≤ N
1 ≤ j1,j2 ≤ M.
^ is Bitwise XOR
Given N and M , you have to calculatethe total number of matrices of size N x M which have both the properties
mentioned above.
Input format:
First line contains T, the number of test cases. 2*T lines follow with N on the first line and M on the second, representing the number of rows and columns respectively.
Output format:
Output the total number of such matrices of size N x M. Since, this answer can be large, output it modulo 10^9+7
Constraints:
1 ≤ N,M,T ≤ 1000
SAMPLE INPUT
1
2
2
SAMPLE OUTPUT
4
Explanation
The four possible matrices are:
[1 3] | [2 3] | [1 4] | [2 4]
[4 2] | [4 1] | [3 2] | [3 1]
题意 构造一个矩阵，使得在矩阵中ij位置，值为a[i]，异或值为i^j,求得比它小的异或值的位置值比它小。且保证矩阵中的数唯一，1~n*m； 暴力做法，三维矩阵异或值图 0 3 2 0 填 1 2 3的排列 为 3！ 1填 4 5 2填 6 7 3 0 1 3填 7 8 2 1 0 排列数为 3！*2！*2！*2！ 由于1000^1000 最大为 1023 .所以暴力可求。 算出所有的 A！
#include <bits/stdc++.h> using namespace std; const int maxn=2333; const int mod=1e9+7; const int mm=1e6+7; int n,m; int cnt[maxn]; int fact[mm]; int main() { fact[0]=1; for(long long i=1;i<=1e6;i++) fact[i]=i*fact[i-1]%mod; int T; scanf("%d",&T); while(T--) { memset(cnt,0,sizeof(cnt)); scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) cnt[i^j]++; } long long ans=1; for(int i=0;i<1024;i++) { if(cnt[i]) { ans=ans*fact[cnt[i]]%mod; } } printf("%lld\n",ans ); } }