Count the number of n×m matrices A satisfying the following condition modulo (109+7).
-
Ai,j∈{0,1,2} for all 1≤i≤n,1≤j≤m.
-
Ai,j≤Ai+1,j for all 1≤i<n,1≤j≤m.
-
Ai,j≤Ai,j+1 for all 1≤i≤n,1≤j<m.
Input
The input consists of several test cases and is terminated by end-of-file.
Each test case contains two integers n and m.
- 1≤n,m≤103
- The number of test cases does not exceed 105.
Output
For each test case, print an integer which denotes the result.
Example
Input
1 2
2 2
1000 1000
Output
6
20
540949876
如图,有两条分界线,设起点分别是A,C终点B,D
Lindstrom-Gessel-Viennot lemma 定理:
对于一张无边权的DAG图,给定n个起点和对应的n个终点,这n条不相交路径的方案数为
(行列式) 其中e(a,b)为图上a到b的方案数
原理不理解,直接套用/(ㄒoㄒ)/~~
AB,CD:对于单独的轮廓线,共需要往上走n步,往右走m步。在这n+m步中选择n步向上走,即有C(n+m,n)种方式。
假设将其中一条轮廓线向左上平移一个单位,有原先的(n,0)->(0,m)变成了(n-1,-1)->(-1,m-1)
相交的情况可以理解成从A到D,从C到B。情况数是C(n+m,n-1)*C(n+m,m-1)
该图片来自lhy同学的友情提供:根据公式:C(n+m,n)*C(n+m,n)-C(n+m,n-1)*C(n+m,m-1)
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll c[2010][2010];
int main()
{
int n,m;
c[1][0]=c[1][1]=1;
for(int i=2;i<=2000;i++)
{
c[i][0]=1;
for(int j=1;j<=2000;j++)
c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
}
while(~scanf("%d%d",&n,&m))
{
ll ans=(c[n+m][n]*c[n+m][n]%mod-c[n+m][n-1]*c[n+m][m-1]%mod+mod)%mod;
printf("%lld\n",ans);
}
}