有 n 个数对 ((A_i); (B_i))，求出
[sum_{i=1}^{n}sum_{j=i + 1}^{n}{a_i+b_i+a_j+b_j choose a_i+a_j}
]
答案对1e9+7取模
- (2le Nle200,000)
- (1le A_ile2000, 1le B_ile2000)
考虑({a_i+b_i+a_j+b_j choose a_i+a_j})的组合意义，相当于从((0,0))开始，每次只能向右或上走一格走到((a_i+a_j,b_i+b_j))的方案数。
于是我们考虑继续把这个限制变成从((-a_i,-b_i))走到((a_j,b_j))的方案数，然后设(f_{i,j})表示从((-M,-M))走到((i,j))的方案数，那么可以得到转移方程：
[egin{cases}f_{-a_i,-b_i}+=1,(1le ile n)\f_{i,j}+=f_{i-1,j-1}end{cases}
]
那么我们统计(sum_{i=1}^nf_{i,j})，然后会发现多算了从((-a_i,-b_i))走到((a_i,b_i))的方案，减去(sum_{i=1}^n{2(a_i+b_i)choose 2a_i})就可以了。
Code
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int N = 2e5;
const int M = 2e3;
const int p = 1e9 + 7;
using namespace std;
int n,a[N + 5],b[N + 5],f[M * 2 + 1][M * 2 + 1],fac[N + 5],inv[N + 5],ans;
int C(int n,int m)
{
return 1ll * fac[n] * inv[m] % p * inv[n - m] % p;
}
int main()
{
scanf("%d",&n);
for (int i = 1;i <= n;i++)
scanf("%d%d",&a[i],&b[i]);
fac[0] = 1;
for (int i = 1;i <= N;i++)
fac[i] = 1ll * fac[i - 1] * i % p;
inv[1] = 1;
for (int i = 2;i <= N;i++)
inv[i] = 1ll * (p - p / i) * inv[p % i] % p;
inv[0] = 1;
for (int i = 1;i <= N;i++)
inv[i] = 1ll * inv[i - 1] * inv[i] % p;
for (int i = 1;i <= n;i++)
f[-a[i] + M][-b[i] + M]++;
for (int i = 1;i <= M * 2;i++)
f[0][i] += f[0][i - 1],f[0][i] %= p;
for (int i = 1;i <= M * 2;i++)
{
f[i][0] += f[i - 1][0];
f[i][0] %= p;
for (int j = 1;j <= M * 2;j++)
f[i][j] += (f[i - 1][j] + f[i][j - 1]) % p,f[i][j] %= p;
}
for (int i = 1;i <= n;i++)
ans += (f[a[i] + M][b[i] + M] - C((a[i] + b[i]) * 2,a[i] * 2)) % p,ans %= p;
ans = 1ll * ans * inv[2] % p;
cout<<(ans + p) % p<<endl;
return 0;
}