Brian Warehouse

You are given natural numbers a and b. Find a^b-b^a.

Input contains numbers a and b (1≤a,b≤100).

Write answer to output.

Sample Input

2 3

Sample Output

-1

一看到这种题目，就想到高精度算法，Java的大数。
此举确实流氓了一点，不过确实过了，鄙人的第一题SGU啊，不许打击。
过段时间看看能不能写出 (C++ Edition) ,啥也别说了，上代码。

import java.math.*;
import java.util.*;

public class Solution
{
   public static void main(String[] args)
   {
      Scanner in = new Scanner(System.in);
      int a = in.nextInt();
      int b = in.nextInt();
      
      BigInteger A=BigInteger.valueOf(a);
      BigInteger B=BigInteger.valueOf(b); // A^B
      BigInteger rA=BigInteger.valueOf(a);
      BigInteger rB=BigInteger.valueOf(b); // store the result after computing
      
      for(int i=1; i<b; i++) // just b-1 times
          rA=rA.multiply(A);
      for(int i=1; i<a; i++)    
          rB=rB.multiply(B);
      System.out.println(rA.subtract(rB)); // sub
   }
}

附《Core Java I》里关于 大数的简单介绍，讲的算比较清晰的了：
Big Numbers
If the precision of the basic integer and floating-point types is not sufficient, you can
turn to a couple of handy classes in the java.math package: BigInteger and BigDecimal. These
are classes for manipulating numbers with an arbitrarily long sequence of digits. The
BigInteger class implements arbitrary precision integer arithmetic, and BigDecimal does the
same for floating-point numbers.
Use the static valueOf method to turn an ordinary number into a big number:
BigInteger a = BigInteger.valueOf(100);
Unfortunately, you cannot use the familiar mathematical operators such as + and * to
combine big numbers. Instead, you must use methods such as add and multiply in the big
number classes.
BigInteger c = a.add(b); // c = a + b
BigInteger d = c.multiply(b.add(BigInteger.valueOf(2))); // d = c * (b + 2)
C++ NOTE: Unlike C++, Java has no programmable operator overloading. There was no way
for the programmer of the BigInteger class to redefine the + and * operators to give the add and
multiply operations of the BigInteger classes. The language designers did overload the + operator
to denote concatenation of strings. They chose not to overload other operators, and they
did not give Java programmers the opportunity to overload operators in their own classes.
Listing 3–6 shows a modification of the lottery odds program of Listing 3–5, updated to
work with big numbers. For example, if you are invited to participate in a lottery in
which you need to pick 60 numbers out of a possible 490 numbers, then this program
will tell you that your odds are 1 in 7163958434619955574151162225400929334117176
12789263493493351 013459481104668848. Good luck!
The program in Listing 3–5 computed the statement
lotteryOdds = lotteryOdds * (n - i + 1) / i;
When big numbers are used, the equivalent statement becomes
lotteryOdds = lotteryOdds.multiply(BigInteger.valueOf(n - i + 1)).divide(BigInteger.valueOf(i));
Listing 3–6 BigIntegerTest.java
1. import java.math.*;
2. import java.util.*;
3.
4. /**
5. * This program uses big numbers to compute the odds of winning the grand prize in a lottery.
6. * @version 1.20 2004-02-10
7. * @author Cay Horstmann
8. */
9. public class BigIntegerTest
10. {
11. public static void main(String[] args)
12. {
13. Scanner in = new Scanner(System.in);
14.
15. System.out.print("How many numbers do you need to draw? ");
16. int k = in.nextInt();
17.
18. System.out.print("What is the highest number you can draw? ");
19. int n = in.nextInt();
20.
21. /*
22. * compute binomial coefficient n*(n-1)*(n-2)*...*(n-k+1)/(1*2*3*...*k)
23. */
24.
25. BigInteger lotteryOdds = BigInteger.valueOf(1);
26.
27. for (int i = 1; i <= k; i++)
28. lotteryOdds = lotteryOdds.multiply(BigInteger.valueOf(n - i + 1)).divide(
29. BigInteger.valueOf(i));
30.
31. System.out.println("Your odds are 1 in " + lotteryOdds + ". Good luck!");
32. }
33. }

java.math.BigInteger

? BigInteger add(BigInteger other)
? BigInteger subtract(BigInteger other)
? BigInteger multiply(BigInteger other)
? BigInteger divide(BigInteger other)
? BigInteger mod(BigInteger other)
returns the sum, difference, product, quotient, and remainder of this big integer and
other.
? int compareTo(BigInteger other)
returns 0 if this big integer equals other, a negative result if this big integer is less
than other, and a positive result otherwise.
? static BigInteger valueOf(long x)
returns a big integer whose value equals x.
? BigDecimal add(BigDecimal other)
? BigDecimal subtract(BigDecimal other)
? BigDecimal multiply(BigDecimal other)
? BigDecimal divide(BigDecimal other, RoundingMode mode) 5.0
returns the sum, difference, product, or quotient of this big decimal and other.
To compute the quotient, you must supply a rounding mode. The mode
RoundingMode.HALF_UP is the rounding mode that you learned in school (i.e., round
down digits 0 . . . 4, round up digits 5 . . . 9). It is appropriate for routine
calculations. See the API documentation for other rounding modes.
? int compareTo(BigDecimal other)
returns 0 if this big decimal equals other, a negative result if this big decimal is less
than other, and a positive result otherwise.
? static BigDecimal valueOf(long x)
? static BigDecimal valueOf(long x, int scale)
returns a big decimal whose value equals x or x /10scale.