Hdoj 2050. Polyline Split plane

Problem Description

We've seen a lot of straight-line split-plane problems, and this topic is slightly different today, and we're asking for the maximum number of n-polyline split planes. For example, a polyline can divide the plane into two parts, and a maximum of two polylines can be divided into 7 parts, as shown below.

Input

The first line of the input data is an integer c, representing the number of test instances, followed by the C row of data, each line containing an integer n (0<n<=10000), representing the number of polylines.

Output

For each test instance, output the maximum number of partitions for the plane, and the output for each instance takes up one row.

Sample Input

212

Sample Output

27

Author

Lcy

Source

Recursive solution topic exercise (for beginner)

Ideas

First consider the case of a straight line, the nth line produces a maximum of \ (n-1\) intersection, the portion of the split is increased by \ (() (N-1) +1 = n\)

Now look at the polyline can be understood as 2 straight lines:

• One and the other part of the polyline ( (N-1) \), is \ (N-1) \), the polyline has 2 lines, that is, the final answer is up to \ (2*2* (N-1) \) the intersection, more increase \ (2*2* (N-1) +1\) Parts

So the recursive type is:\ (F[i] = f[i-1] + (i-1) + 1;\)

Initial conditions:\ (f[1]=2\)

Code

#include<bits/stdc++.h>using namespace std;__int64 f[10010];int main(){    int t;        f[1] = 2; f[2] = 7;    for(int i=3;i<=10000;i++)        f[i] = f[i-1] + 4*(i-1) + 1;        cin >> t;    while(t--)    {        int tmp;        cin >> tmp;         cout << f[tmp] << endl;    }    return 0;   }

Hdoj 2050. Polyline Split plane