hdu 5621 KK's Point【思维】

KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 269    Accepted Submission(s): 93

Problem Description

Our lovely KK has a difficult mathematical problem:He pointsN(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

234

 

Sample Output

35

题目大意：

在一个圆上点下n个点，问在圆内能够组成多少个交点（圆上的点也包括）、这里思路很好形成：

在圆上点下1个点的时候输出1

在圆上点下2个点的时候输出2

在圆上点下3个点的时候输出3

在圆上点下4个点的时候输出5

因为呢，只有点上4个点的时候才能在园内出现一个交点，

那么在圆上点下5个点的时候呢？思路很简单：从5个点中不断的选出4个点就能够组成一个交点，那么就是CN4、

所以最终结果就是CN-4 4

这里直接上代码：

#include<stdio.h>#include<string.h>using namespace std;#define llu unsigned long long int//数据较大，ll会wa、int main(){    int t;    scanf("%d",&t);    while(t--)    {        llu n;        scanf("%I64u",&n);        llu output=n;        if(n==0||n==1||n==2||n==3)        {            printf("%I64u\n",output);            continue;        }        else        {            output+=n*(n-1)/2*(n-2)/3*(n-3)/4;            printf("%I64u\n",output);        }    }    return 0;}