hdoj--5621--KK's Point（简单数学）

KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 350    Accepted Submission(s): 121

Problem Description

Our lovely KK has a difficult mathematical problem:He points
N(2≤N≤105)
points on a circle,there are all different.Now he's going to connect the
N
points with each other(There are no three lines in the circle to hand over a point.).KK wants to know how many points are there in the picture(Including the dots of boundary).

 

Input

The first line of the input file contains an integer
T(1≤T≤10),
which indicates the number of test cases.



For each test case, there are one lines,includes a integer
N(2≤N≤105),indicating
the number of dots of the polygon.

 

Output

For each test case, there are one lines,includes a integer,indicating the number of the dots.

 

Sample Input

2
3
4

 

Sample Output

3
5

 

Source

BestCoder Round #71 (div.2)

 

有交点需要有3个或者4个点，三个点的话，交点就是有公共点，相交的更像是一个角，四个点的话就是两天直线相交，因为题中说两两相交，那说明每一条线都会跟其他的线相交，并且不会有重复的交点，显然就是组合数C（4,4）+n，但是数据范围比较大，所以需要unsigned long long

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;
int main()
{
	int t;
	cin>>t;
    while(t--)
    {
		unsigned long long n;
	 	cin>>n;
	  	unsigned long long temp=n*(n-1)*(n-2)/6*(n-3)/4+n;
	    cout<<temp<<endl;
    }
	return 0;
}