题目链接：

题意：给定n个三角形，问最多可以把区域化成多少个部分，这是一个一维空间  一定会满足一元二次方程  题目给定1 2的个数 只要得到3的个数就可以用待定系数法求得公式：F(x) = 3*(x-1)*x+2;  另外如果是二维的话，会满足一元三次方程 ，也可以用待定系数法求解；20

AC代码：

1 import java.io.BufferedReader; 2 import java.io.IOException; 3 import java.io.InputStream; 4 import java.io.InputStreamReader; 5 import java.io.PrintWriter; 6 import java.util.StringTokenizer; 7  8 public class Main { 9     public static void main(String[] args) {10         InputReader s = new InputReader(System.in);11         PrintWriter cout = new PrintWriter(System.out);12         int t , x,t1;13         t  =   s.nextInt();14         while (t-- > 0) {15             x = s.nextInt();16              t1 = 3*(x-1)*x+2;17             cout.println(t1);18          19         }20         cout.flush();21     }22     static int gcd(int a, int b) {23         return b == 0 ? a : gcd(b, a % b);24     }25 }26 class InputReader {27 28     public BufferedReader rea;29     public StringTokenizer tok;30 31     public InputReader(InputStream stream) {32         rea = new BufferedReader(new InputStreamReader(stream), 32768);33         tok = null;34     }35 36     public String next() {37         while (tok == null || !tok.hasMoreTokens()) {38             try {39                 tok = new StringTokenizer(rea.readLine());40             } catch (IOException e) {41                 throw new RuntimeException(e);42             }43         }44         return tok.nextToken();45     }46 47     public int nextInt() {48         return Integer.parseInt(next());49     }50 51 }