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. 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]
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：找出有多少个排列矩阵 使得下标i1^j1>j2^j2 对应的元素也为a[i1][j1]>a[i2][j2]
排列矩阵是包含1-n*m每个数的矩阵
题解：先考虑一个矩阵n*m的答案，我们只要把i^j相同的位置做排列就行了。因为他们两个的大小没法比，所以可以排列。
因为i^j<=1024 所以记录数字的多少开1024即可。
然后考虑t个n*m
令n<m
我们可以处理出行的前缀和
如果m<=500，那么我们暴力 复杂度是500*500
如果m>500 那么我们取前n行的前缀和，减去前n行 m+1到1000列的数即可 复杂度500*500
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<deque> using namespace std; typedef long long ll; ll facts[1200],num[1050],sum[1005][1050],vis[1050]; int main(){ ll i,j,t,n,m; facts[0]=1; for(i=1;i<=1024;i++)facts[i]=facts[i-1]*i00000007; for(i=1;i<=1000;i++){ for(j=1;j<=1000;j++){ num[i^j]++; sum[i][i^j]++; } for(j=0;j<=1024;j++)sum[i][j]+=sum[i-1][j]; } scanf("%lld",&t); while(t--){ scanf("%lld%lld",&n,&m); if(n>m)swap(n,m); memset(vis,0,sizeof(vis)); ll ans=1; if(m<=500){ for(i=1;i<=n;i++){ for(j=1;j<=m;j++){ vis[i^j]++; } } } else{ for(i=0;i<=1024;i++)vis[i]=sum[n][i]; for(i=1;i<=n;i++){ for(j=m+1;j<=1000;j++){ vis[i^j]--; } } } for(i=0;i<=1024;i++)ans=ans*facts[vis[i]]00000007; printf("%lld\n",ans); } return 0; }
