KK's Point

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 300    Accepted Submission(s): 103

Problem Description

N(2≤N≤105) points on a circle,there are all different.Now he's going to connect the  N

Input

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)

这个题在比赛的时候没想到题解的思路，看了题解好简单，

两个方法:

方法1  找规律：

其实随便画几个就能看出规律来，

我直接说出规律了，，比如说N = 8的时候，

那么结果就是：

S1 = 1 + 2 + 3 + 4 + 5

S2 = 2 + 4 + 6 + 8

S3 = 3 + 6 + 9

S4 = 4 + 8

S5 = 5

S1 + S2 +... + S5 在加上N，就是结果！

规律很明显，很容易敲出代码了！

#include<iostream>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<string>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<cctype>
#include<cstdlib>
#define mem(x) memset(x,0,sizeof(x));
#define mem1(x) memset(x,-1,sizeof(x));
using namespace std;
const int maxn = 1000 + 10;
const int maxt = 75 + 10;
const double eps = 1e-8;
const int INF = 1e8;
const double pi = acos(-1.0);
typedef long long ll;
typedef unsigned long long llu;
int main()
{
    int n,T;
    cin >> T;
    while(T--){
        cin >> n;
        llu num = n-3;
        llu sum = n;
        for (int i = 1; i <= n-2; ++i){
            sum += num*i+num*(num-1)/2*i;
            num--;
        }
        cout << sum << endl;
    }
    return 0;
}

方法2 ：（题解方法）

先不考虑在圆上的点，一共有N个点，那么里面有几个四边形，就有几个交点，比如说N = 4，就只能构成一个四边形，所以里面有1个交点加上圆上4个正好5个.

为什么这么想可以呢，因为题目保证不会有三边交于一点，那么不就是四边形的对角线的交点吗，所以说有几个四边形就有几个交点。

所以答案是C N 4 + N;