BestCoder Round #71 (div.2) （hdu 5621）

KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 644    Accepted Submission(s): 220

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

题意：一个圆上给n个点，将这些点两两相连，问有多少个交点

要求：1、圆上的点也算   2、圆内只有两条线的交点，没有两条以上线的交点

题解：圆上的点为n个，我们只需计算圆内有多少个交点即可，4个点可以在圆内形成一个交点，所以我们计算n个点可以组成多少个四边形即可

求解组合公式C(n,4)+n;

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAX 100100
#define INF 0x3f3f3f
using namespace std;
int main()
{
	int t;
	unsigned long long n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%llu",&n);
		if(n<4)
		{
			printf("%llu\n",n);
			continue;
		}
		unsigned long long ans=n*(n-1)/2*(n-2)/3*(n-3)/4+n;
		printf("%llu\n",ans);
	}
	return 0;
}