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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

AGC001E BBQ Hard

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LuoGu: AT1983 [AGC001E] BBQ Hard

AtCoder: E - BBQ Hard

考虑计算 $\displaystyle \sum _ {i = 1} ^ n \sum _ {j = 1} ^ n f _ {i, j}$ ，将自己对自己的贡献减去后除以 $2$ 即为答案。

考虑组合意义，$\binom {a _ i + b _ i + a _ j + b _ j} {a _ i + a _ j}$ 表示从 $(0, 0)$ 到 $(a _ i + a _ j, b _ i + b _ j)$ 只向上或右走的方案数，平移，也就是 $(-a _ j, -b _ j)$ 到 $(a _ i, b _ i)$ 的方案数。

这样 $O(m ^ 2)$ DP 算而不是用数学算可以对于每一个元素 $O(1)$ 获得所有元素对他的贡献。

查看代码 
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#include <cstdio>
using namespace std;
template <class Type>
void read(Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 2e5 + 10, M = 2e3 + 10, mod = 1e9 + 7;
void adj (int &x) { x += x >> 31 & mod; };
int binpow (int b, int k)
{
    int res = 1;
    for (; k; b = (LL)b * b % mod, k >>= 1)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
int n, A[N], B[N], f[M << 1][M << 1];
int inv[M << 2], fact[M << 2], ifact[M << 2];
void init()
{
    inv[1] = 1;
    for (int i = 2; i < M << 2; i++)
        inv[i] = (LL)(mod - mod / i) * inv[mod % i] % mod;
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < M << 2; i++)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
    }
}
int C (int a, int b) { return (LL)fact[a] * ifact[a - b] % mod * ifact[b] % mod; }
int main ()
{
    read(n);
    for (int i = 1; i <= n; ++i)
    {
        read(A[i], B[i]);
        f[M - A[i]][M - B[i]]++;
    }
    init();
    for (int i = 1; i < M << 1; ++i)
        for (int j = 1; j < M << 1; ++j)
            adj(f[i][j] += f[i - 1][j] - mod), adj(f[i][j] += f[i][j - 1] - mod);
    int res = 0;
    for (int i = 1; i <= n; i++)
    {
        adj(res += f[M + A[i]][M + B[i]] - mod);
        adj(res -= C((A[i] << 1) + (B[i] << 1), A[i] << 1));
    }
    adj(res = (LL)res * inv[2] % mod);
    write(res);
    return 0;
}
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