2016huasacm暑假集训训练四 递推_B

题目链接：https://vjudge.net/contest/125308#problem/B

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：

 import java.io.BufferedReader;

 import java.io.IOException;

 import java.io.InputStream;

 import java.io.InputStreamReader;

 import java.io.PrintWriter;

 import java.util.StringTokenizer;

 public class Main {

     public static void main(String[] args) {

         InputReader s = new InputReader(System.in);

         PrintWriter cout = new PrintWriter(System.out);

         int t , x,t1;

         t  =   s.nextInt();

         while (t-- > ) {

             x = s.nextInt();

              t1 = *(x-)*x+;

             cout.println(t1);

         }

         cout.flush();

     }

     static int gcd(int a, int b) {

         return b ==  ? a : gcd(b, a % b);

     }

 }

 class InputReader {

     public BufferedReader rea;

     public StringTokenizer tok;

     public InputReader(InputStream stream) {

         rea = new BufferedReader(new InputStreamReader(stream), );

         tok = null;

     }

     public String next() {

         while (tok == null || !tok.hasMoreTokens()) {

             try {

                 tok = new StringTokenizer(rea.readLine());

             } catch (IOException e) {

                 throw new RuntimeException(e);

             }

         }

         return tok.nextToken();

     }

     public int nextInt() {

         return Integer.parseInt(next());

     }

 }