SWJTU2017-6月月赛 G-A Easy Counting Problem[数论][乘法逆元]

传送门：http://www.swjtuoj.cn/problem/2397/

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题解：产生交点的条件为4个点构成四边形对角线产生交点，最大解当产生的交点位置完全不相同时存在。答案为$C_{\text{n}}^4$

计算组合数时需要使用乘法逆元

代码：

#define _CRT_SECURE_NO_DEPRECATE
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>  
#include<cstdio>  
#include<fstream>  
#include<iomanip>
#include<algorithm>  
#include<cmath>  
#include<deque>  
#include<vector>  
#include<assert.h>
#include<bitset>
#include<queue>  
#include<string>  
#include<cstring>  
#include<map>  
#include<stack>  
#include<set>
#include<functional>
#define pii pair<int, int>
#define mod 1000000007
#define mp make_pair
#define pi acos(-1)
#define eps 0.00000001
#define mst(a,i) memset(a,i,sizeof(a))
#define all(n) n.begin(),n.end()
#define lson(x) ((x<<1))  
#define rson(x) ((x<<1)|1) 
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e4 + 5;
ll poww(ll m, int n)
{
    ll ans = 1;
    ll temp = m;
    while (n)
    {
        if (n & 1)
            ans *= temp;
        temp *= temp;
        ans %= mod;
        temp %= mod;
        n >>= 1;
    }
    return ans%mod;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int T;
    ll n;
    cin >> T;
    while (T--)
    {
        cin >> n;
        ll temp = n;
        for (ll i = 1; i <= 3; ++i)
        {
            ll tt = poww(i + 1, mod - 2);
            temp = (temp*(n - i)) % mod;
            temp = (temp*tt) % mod;
        }
        cout << temp << endl;
    }
    return 0;
}