題意：Farmer John owns a farm. He first builds a circle fence. Then, he will choose n points and build some straight fences connecting them. Next, he will feed a cow in each region so that cows cannot play with each other without breaking the fences. In order to feed more cows, he also wants to have as many regions as possible. However, he is busy building fences now, so he needs your help to determine what is the maximum number of cows he can feed if he chooses these n points properly.

思路：

尤拉定理：f = e - v + 2

v = n + C（n，4）（每4個點的連線必有一個交點）

e = n（圓弧） + C（n，2）+ 2*C（n，4）（一個交點是兩條邊形成的，這個交點將這兩條邊劃分成4條邊，數量上+2，因此有多少交點就增加交點數*2條邊）

ans = f - 1,圓外的不算

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;

LL power_mod(LL a,LL b)
{
    LL base=a%mod;
    a=a%mod;
    LL res=1LL;
    while(b)
    {
        if(b&1)
            res=base*res%mod;
        base=base*base%mod;
        b=b>>1;
    }
    return res%mod;
}

LL C(LL n, LL m)
{
    if(m > n) return 0;
    LL ans = 1;
    for(int i=1; i<=m; i++)
    {
        LL a = (n + i - m) % mod;
        LL b = i % mod;
        ans = ans * (a * power_mod(b, mod-2) % mod) % mod;
    }
    return ans;
}

LL Lucas(LL n, LL m)
{
    if(m == 0) return 1;
    return C(n % mod, m % mod) * Lucas(n / mod, m / mod) % mod;
}

int main() {
    int T;
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++) {
        LL n;
        scanf("%lld", &n);
        printf("Case #%d: %lld\n", kase, (Lucas(n, 2) + Lucas(n, 4) + 1) % mod);
    }
    return 0;
}