hdoj 5621 KK's Point 【数学】

KK's Point

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

Total Submission(s): 478    Accepted Submission(s): 164

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

234

 

Sample Output

35

 

题意：在圆上有n个点，两两相连，保证任意三条线不会交于一点，问有多少交点。

思路：看有多少个四边形，已经保证了任意三条线不会交于一点。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 1000000#define eps 1e-8#define MAXN (10000+10)#define MAXM (10000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while((a)--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)#pragma comment(linker, "/STACK:102400000,102400000")#define fi first#define se secondusing namespace std;typedef pair<int, int> pii;int main(){    int t; Ri(t);    W(t)    {        unsigned long long n; scanf("%I64u", &n);        if(n < 4)        {            cout << n << endl;            continue;        }//        10//        100000//        4166416671250075000        unsigned long long ans = n * (n-1) / 2 * (n-2) / 3 * (n-3) / 4 + n;        cout << ans << endl;    }    return 0;}