题目链接：

KK‘s Point

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Our lovely KK has a difficult mathematical problem:He points N(2≤N≤10^5) 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≤10^5),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,每两个点都相连,问连线有多少个交点,任意三根线不交于同一点;

思路：

dp来解决,每加入一个点我们可以发现,增加的交点个数为1*n+2*(n-1)+3*(n-2)+...+n*1+1(最后这个1是这个点本身);

这是怎么来的可以画个图来自己画一下看看;每次把原先的点分成两拨;

现在的问题变成怎么求上述的式子了;

f(n)=1*n+2*(n-1)+3*(n-2)+...+n*1;

f(n+1)=1*(n+1)+2*n+3*(n-1)+...+(n+1)*1;

f(n+1)-f(n)=(n+1)+n+(n-1)+(n-2)+...+1=(n+1)(n+2)/2;

那么状态转移方程就有了dp[i]=dp[i-1]+1+f(i-3);

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll inf=1e15;
const int N=1e5+6;
int n;
ll dp[N],f[N];
void fun()
{
    dp[1]=1;//注意dp[1]   dp[2]的初始化,我在这里就wa了一发;
    dp[2]=2;
    dp[3]=3;
    f[1]=1;
    for(int i=2;i<N;i++)
    {
        ll x=(i);
        f[i]=f[i-1]+x*(x+1)/2;
    }
    for(int i=4;i<N;i++)
    {
        dp[i]=dp[i-1]+1+f[i-3];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    fun();
    while(t--)
    {
        scanf("%d",&n);
        cout<<dp[n]<<"\n";
        //printf("%lld\n",dp[n]);
    }
    return 0;
}

hdu-5621 KK's Point(dp+数学)