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());
     }

 }