c/c++语言开发共享agc001E – BBQ Hard(dp 组合数)

题意

sol

非常妙的一道题目。

首先，我们可以把(c_{a_i + b_i + a_j + b_j}^{a_i + a_j})看做从((-a_i, -b_i))走到((a_j, b_j))的方案数

然后全都放的一起dp，(f[i][j])表示从((i, j))之前的所有点到((i, j))的方案数

减去重复的即可

 #include<bits/stdc++.h> using namespace std; const int maxn = 2e5 + 10, mod = 1e9 +  7; inline int read() {     int x = 0, f = 1; char c = getchar();     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, a[maxn], b[maxn], f[5001][5001], fac[10001], ifac[10001]; int add(int x, int y) {     if(x + y < 0) return x + y + mod;     return x + y > mod ? x + y - mod : x + y ; } int mul(int x, int y) {     return 1ll * x * y % mod; } int fastpow(int a, int p) {     int base = 1;     while(p) {         if(p & 1) base = mul(base, a);         a = mul(a, a); p >>= 1;     }     return base; } void init() {     fac[0] = 1;     for(int i = 1; i <= 8000; i++) fac[i] = mul(i, fac[i - 1]);     ifac[8000] = fastpow(fac[8000], mod - 2);     for(int i = 8000; i; i--) ifac[i - 1] = mul(i, ifac[i]); } int id(int x) {     return 2001 + x; } int c(int n, int m) {     return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod; } main() { //  freopen("a.in", "r", stdin);     init();     n = read();     for(int i = 1; i <= n; i++) a[i] = read(), b[i] = read(), f[id(-a[i])][id(-b[i])]++;     for(int i = 1; i <= 4221; i++)         for(int j = 1; j <= 4221; j++)                 f[i][j] = add(f[i][j], add(f[i - 1][j], f[i][j - 1]));             //  printf("%d %d %dn", i, j, f[i][j]);     int sum = 0;     for(int i = 1; i <= n; i++)          sum = add(sum, add(f[id(a[i])][id(b[i])], -c(a[i] + b[i] + a[i] + b[i], a[i] + a[i])));         //这里会到8000.。。      sum = 1ll * sum * 500000004ll % mod;     cout << sum % mod;     return 0; } /* 8 2000 2000 1999 1998 1 1 1 1 2 1 1 3 2 1 3 3 */