atcoderAGC001 E

E - BBQ Hard
Time limit : 2sec / Memory limit : 256MB
Score : 1400 points
Problem Statement
Snuke is having another barbeque party.
This time, he will make one serving of Skewer Meal.
He has a stock of N Skewer Meal Packs. The i-th Skewer Meal Pack contains one skewer, Ai pieces of beef and Bi pieces of green pepper. All skewers in these packs are different and distinguishable, while all pieces of beef and all pieces of green pepper are, respectively, indistinguishable.
To make a Skewer Meal, he chooses two of his Skewer Meal Packs, and takes out all of the contents from the chosen packs, that is, two skewers and some pieces of beef or green pepper. (Remaining Skewer Meal Packs will not be used.) Then, all those pieces of food are threaded onto both skewers, one by one, in any order.
(See the image in the Sample section for better understanding.)
In how many different ways can he make a Skewer Meal? Two ways of making a Skewer Meal is different if and only if the sets of the used skewers are different, or the orders of the pieces of food are different. Since this number can be extremely large, find it modulo 109+7.
Constraints
2≦N≦200,000
1≦Ai≦2000,1≦Bi≦2000
Input
The input is given from Standard Input in the following format:
N
A1 B1
A2 B2
:
AN BN
Output
Print the number of the different ways Snuke can make a serving of Skewer Meal, modulo 109+7.
Sample Input 1
Copy
3
1 1
1 1
2 1
Sample Output 1
Copy
26
The 26 ways of making a Skewer Meal are shown below. Gray bars represent skewers, each with a number denoting the Skewer Meal Set that contained the skewer. Brown and green rectangles represent pieces of beef and green pepper, respectively.

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对于两组$(i,j)$
我们可以得到贡献为$C_{a_i+a_j+b_i+b_j}^{a_i+b_i}$
直接求不好求
我们发现$C_{a_i+a_j+b_i+b_j}^{a_i+b_i}$其实就是点$(-a_i,-b_i)$到点$(a_j,b_j)$的方案数
然后我们就可以得到答案的式子
$$(\sum _{i=1}^n\sum _{j=1}^n{C}_{a_i+a_j+b_i+b_j}^{a_i+a_j}-\sum C_{2\ast (a_i+b_i)}^{2\ast a_i})/2$$
后面那个$\sum$很好求，前面那个$\sum$含义就是第三象限的点到第一象限的方案数。直接$dp[i][j]=dp[i-1][j]+dp[i][j-1]$即可

#include
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair
#define Pil pair
#define Pli pair
#define Pll pair
#define pb push_back 
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
int rd()
{
    int sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48,c = getchar();
    if(flag) return sum;
    else return -sum;
}
const int p = 1e9+7;
int n;
int a[201000],b[201000];
int fac[101000],inv[101000];
int det = 2001;
int f[4010][4010];
int C(int i,int j)
{
    int now = 1ll*fac[i]*inv[j]%p;
    now = 1ll*now*inv[i-j]%p;
    return now;
}
int Pow(int a,int x)
{
    int now = 1;
    while(x)
    {
        if(x&1) now = 1ll*now*a%p;
        a = 1ll*a*a%p;
        x >>= 1;
    }
    return now;
}
int main()
{
    n = rd();
    rep(i,1,n) a[i] = rd(),b[i] = rd();
    fac[0] = 1;
    rep(i,1,10000) fac[i] = 1ll*fac[i-1]*i%p,inv[i] = Pow(fac[i],p-2);
    rep(i,1,n) f[det-a[i]][det-b[i]] += 1;
    int ans = 0;
    rep(i,1,4001) rep(j,1,4001) (f[i][j] += (f[i][j-1]+f[i-1][j])%p)%=p;
    rep(i,1,n) (ans += f[a[i]+det][b[i]+det])%=p;
    rep(i,1,n) ans = ((ans - C(2*a[i]+2*b[i],2*a[i]))%p+p)%p;
    ans = 1ll*ans*Pow(2,p-2)%p;
    printf("%d\n",ans);
    return 0;
}