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-- > 0) {
            x = s.nextInt();
             t1 = 3*(x-1)*x+2;
            cout.println(t1);
         
        }
        cout.flush();
    }
    static int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}
class InputReader {

    public BufferedReader rea;
    public StringTokenizer tok;

    public InputReader(InputStream stream) {
        rea = new BufferedReader(new InputStreamReader(stream), 32768);
        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());
    }

}