[DP]AGC001E

Description

求

∑i=1n∑j=i+1n(Ai+Aj+Bi+BjAi+Aj)

n≤2×105,Ai,Bi≤2×103

Solution

可能可以二维FFT？？？
这个组合数就是从(−Ai,−Bi)到(Aj,Bj)的路径条数。
那直接O(A2i)DP就好了。

#include <bits/stdc++.h>using namespace std;const int N = 202020;const int M = 4040;const int MOD = 1000000007;const int s = 2020;typedef long long ll;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}template<typename T>inline void read(T &x) {    static char c; x = 0; int sgn = 0;    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';    if (sgn) x = -x;}int a[N], b[N];int f[M][M];int fac[N], inv[N];int n, m, ans;inline void Add(int &x, int a) {    x += a; while (x >= MOD) x -= MOD;}inline int C(int n, int m) {    return (ll)fac[n + m] * inv[n] % MOD * inv[m] % MOD;}int main(void) {    freopen("1.in", "r", stdin);    read(n);    for (int i = 1; i <= n; i++) {        read(a[i]); read(b[i]);        ++f[s - a[i]][s - b[i]];    }    for (int i = 1; i < M; i++)        for (int j = 1; j < M; j++) {            Add(f[i][j], f[i - 1][j]);            Add(f[i][j], f[i][j - 1]);        }    inv[1] = 1;    for (int i = 2; i <= 10000; i++)        inv[i] = (ll)(MOD - MOD / i) * inv[MOD % i] % MOD;    fac[0] = inv[0] = 1;    for (int i = 1; i <= 10000; i++) {        fac[i] = (ll)i * fac[i - 1] % MOD;        inv[i] = (ll)inv[i] * inv[i - 1] % MOD;    }    for (int i = 1; i <= n; i++) {        Add(ans, f[s + a[i]][s + b[i]]);        Add(ans, MOD - C(a[i] << 1, b[i] << 1));    }    ans = (ll)ans * (MOD + 1) / 2 % MOD;    cout << ans << endl;    return 0;}