HDOJ2116 Has the sum exceeded

Has the sum exceeded

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4869    Accepted Submission(s): 1093

Problem Description

As we all know, in the computer science, an integer A is in the range of 32-signed integer, which means the integer A is between -2^31 and (2^31)-1 (inclusive), and A is a 64-signed integer, which means A is between -2^63 and (2^63)-1(inclusive). Now we give the K-signed range, and two K-signed integers A and B, you should check whether the sum of A and B is beyond the range of K-signed integer or not.

Input

There will be many cases to calculate. In each case, there comes the integer K (2<=K<=64) first in a single line. Then following the line, there is another single line which has two K-signed integers A and B.

Output

For each case, you should estimate whether the sum is beyond the range. If exceeded, print “Yes”, otherwise “WaHaHa”.

Sample Input

32 100 100

Sample Output

WaHaHa

判断A+B的和是否在（-2^(k-1)，2^k  -1）范围里

需要打表，并且使用大数来做

import java.math.BigInteger;
import java.util.Scanner;

public class Main{
  private static Scanner scanner;
  private static long arr[];

  public static void main(String[] args) {
    scanner = new Scanner(System.in);
    function() ;
    while (scanner.hasNext()) {
      int n = scanner.nextInt();
      String aString = scanner.next();
      String bString = scanner.next();
      BigInteger aInteger = new BigInteger(aString);
      BigInteger bInteger = new BigInteger(bString);
      aInteger = aInteger.add(bInteger);// 和
      
      BigInteger maxInteger = new BigInteger("" + (arr[n] - 1));// 范围的最大值
      BigInteger minInteger = new BigInteger("" + (-arr[n]));// 范围的最小值
      if (aInteger.compareTo(maxInteger) > 0 || aInteger.compareTo(minInteger) < 0) {
        System.out.println("Yes");
      } else {
        System.out.println("WaHaHa");
      }
    }
  }

  public static void function() {
    arr = new long[65];
    arr[1] = 1;
    for (int i = 2; i <= 64; i++) {
      arr[i] = arr[i - 1] * 2;
    }
  }
}