AT1983 [AGC001E] BBQ Hard

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

• \(2\le N\le200,000\)
• \(1\le A_i\le2000,\ 1\le B_i\le2000\)

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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：

\[\begin{cases}f_{-a_i,-b_i}+=1,(1\le i\le n)\\f_{i,j}+=f_{i-1,j-1}\end{cases} \]

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;
}