hdoj 5621 KK's Point 【数学】

最新推荐文章于 2017-01-19 15:42:22 发布

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最新推荐文章于 2017-01-19 15:42:22 发布

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本篇介绍了一个有趣的数学问题：给定圆周上的N个不同点，两两相连且任意三条线不交于一点，求解形成的交点总数。文章提供了问题的解析思路及AC代码实现。

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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

2
3
4

Sample Output

3
5

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 second
using 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;
}