hdu--5621

KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 442    Accepted Submission(s): 148

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)

 

解体思路:在圆上画出n个点,每个点两两相连, 在圆内没有三条直线相交成的点,但在圆上可以,问包括圆上n个点,一共有多少个交点,这肯定是一个规律题,.当n>=4时,圆内才会出现交点,可以通过画图发现,圆内每一个不同的交点都由一个不同的四边形得到,而四边形的顶点,都来自所给的n个顶点.所以当年n>=4时,交点数为Cn4+n.否则为n.这题中注意取值范围,以unsigned long long int 输出，long long 会wa.

代码如下：

#include<stdio.h>
#include<string.h>
#define llu unsigned long long int
int main()
{
    int t;
    llu m,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%llu",&n);
        if(n==2||n==3){
        	printf("%llu\n",n);
        	continue;
        }
        m=n*(n-1)/2*(n-2)/3*(n-3)/4;//每二个连续数 中一个是偶数,每3个连续数中一个3的倍数..... 
        printf("%llu\n",m+n);
    }
    return 0;
}