题意：http://codeforces.com/problemset/problem/295/D
思路：
可以把图形看成上下两个金字塔（极端情况下长方形），dp出以i的长为底，以j为高的三角形数（长方形），再预处理前缀（以i的长为底，以<=j为高），然后就枚举中心行，上下金字塔数相乘，注意两个图形的底边相同导致重复的可能性
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)2e3+10; 51 52 ll sq[N][N]; 53 ll sum[N]; 54 55 int main() 56 { 57 int n,m; 58 sc("%d%d",&n,&m); 59 if(m<2) 60 { 61 pr("0 "); 62 return 0; 63 } 64 for(int i=2;i<=m;++i) 65 sq[i][1]=1; 66 for(int i=2;i<=n;++i) 67 { 68 sq[2][i]=1; 69 sum[2]=1; 70 for(int j=3;j<=m;++j) 71 sum[j]=sq[j][i-1]+sum[j-1],sum[j]%=mod; 72 ll temp=1; 73 for(int j=3;j<=m;++j) 74 temp+=sum[j],temp%=mod,sq[j][i]=temp; 75 } 76 // for(int i=1;i<=n;++i) 77 // for(int j=2;j<=m;++j) 78 // { 79 // pr("%d %d: %lld ",j,i,sq[j][i]); 80 // } 81 for(int j=2;j<=m;++j) 82 for(int i=2;i<=n;++i) 83 { 84 sq[j][i]+=sq[j][i-1]; 85 sq[j][i]%=mod; 86 } 87 ll ans=0; 88 // for(int i=2;i<=m;++i)sq[i][0]=1; 89 for(int i=1;i<=n;++i) 90 { 91 for(int j=2;j<=m;++j)sum[j]=sq[j][n-i]; 92 for(int j=3;j<=m;++j)sum[j]+=sum[j-1],sum[j]%=mod; 93 ans+=sq[2][i]*(m-2+1)%mod; 94 ans%=mod; 95 ll temp=0; 96 for(int j=3;j<=m;++j) 97 ans+=sq[j][i]*(m-j+1)%mod,ans%=mod; 98 for(int j=3;j<=m;++j) 99 { 100 temp+=sum[j-1]; 101 temp%=mod; 102 temp+=sq[j-1][n-i]; 103 temp%=mod; 104 ans+=sq[j][i]*temp%mod*(m-j+1)%mod; 105 ans%=mod; 106 // temp+=sum[j]; 107 } 108 } 109 pr("%lld ",ans); 110 return 0; 111 } 112 113 /**************************************************************************************/