1010: Drawing Lines

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 2134 Solved: 856
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Description

Little Johnny likes to draw a lot. A few days ago he painted lots of straight lines on his sheet of paper. Then he counted in how many zones the sheet of paper was split by these lines. He noticed that this number is not always the same. For instance, if he draws 2 lines, the sheet of paper could be split into 4, 3 or even 2 (if the lines are identical) zones. Since he is a very curious kid, he would like to know which is the maximum number of zones into which he can split the sheet of paper, if he draws N lines. The sheet of paper is to be considered a very large (=infinite) rectangle.

Input

The input file will contain an integer number:T(T<=1000);

then will contain T integers :N (0<=N<=65535).

Output

You should output: the maximum number of zones into which the sheet of paper can be split if Johnny draws N lines

Sample Input

2
0
1

Sample Output

1
2

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
	int t;
	long long int n;
	scanf("%d",&t);
	while(t--)
    {
        cin>>n;
        n=n*(n+1)/2+1;
        cout<<n<<endl;
    }
	return 0;
}

一，直线分割平面：

首先考虑 n条直线最多把平面分成an部分

于是a0=1 a1=2 a2=4

对于已经有n条直线将平面分成了最多的an块

那么加一条直线他最多与前n条直线有n个交点于是被它穿过的区域都被一分为二那么增加的区域数就是穿过的区域

数也就是这条直线自身被分成的段数就是n+1 故 a(n+1) = an+n+1

an = n+(n-1)+...+2+a1 = n(n+1)/2 +1

二，平面分割空间：

设n个平面最多把空间分成bn个部分

于是b0=1 b1=2 b2=4

对于已经有n个平面将空间分成了最多的bn块

那么加入一个平面它最多与每个平面相交在它的上面就会得到至多n条交线

同时被它穿过的空间区域也被它一分为二那么增加的区域数仍旧是它穿过的区域数也就是这个平面自身被直线分割

成的块数就是an

于是b(n+1)=bn+an

bn=a(n-1)+b(n-1)=...=a(n-1)+a(n-2)+...+a1+b1

=(n-1)n/2 +(n-2)(n-1)/2+...+1*(1+1)/2+n+2

=求和[1方到(n-1)方]/2 + 求和[1到(n-1)]/2 +n+1

=n(n-1)(2n-1)/12 +n(n-1)/4 +n+1

=n(n+1)(n-1)/6 +n+1

=(n^3+5n+6)/6

三，直线分割空间：

由上所述，直接可以得到公式： an = (n*(n+1)/2+1) * (n^3+5n+6)/6

= (n^5+n^4+7*n^3+11*n^2+16*n)/12+1

四，折线分割平面：

根据直线分平面可知，由交点决定了射线和线段的条数，进而决定了新增的区域数。当n-1条折线时，区域数为f（n-1）。为了使增加的区域最多，则折线的两边的线段要和n-1条折线的边，即2*（n-1）条线段相交。那么新增的线段数为4*（n-1），射线数为2。但要注意的是，折线本身相邻的两线段只能增加一个区域。

故：f(n)=f(n-1)+4(n-1)+2-1

=f(n-1)+4(n-1)+1

=f(n-2)+4(n-2)+4(n-1)+2

……

=f(1)+4+4*2+……+4(n-1)+(n-1)

=2n^2-n+1