Input

Input contains integer number N (1<=N<=106)
首先想到的是枚举，我确实也这么做了，5~8无果，对于9，我的05年方正 Pentium M  上跑了足足1分半钟之久，我一开始还以为死循环了，后来出来八个结果，也去网上核对了一下，确实满足的只有这八个:
   111111111
   119357639
   380642361
   388888889
   611111111
   619357639
   880642361
   888888889
我没有学过数论，找到一篇博文，对于数论解法讲的还算清楚，不过用pascal写就，我自己用C写了一下，第三次AC了。博文链接如下：http://blog.csdn.net/Skyprophet/archive/2009/10/05/4634801.aspx

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3. (找出这个数列1-N号元素中能被3整除的有多少个)

Input contains N (1<=N<=2³¹ - 1).

Nearly prime number is an integer positive number for which it is possible to find such primes P₁ and P₂ that given number is equal to P₁*P₂. There is given a sequence on N integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime and “No” otherwise.

Input file consists of N+1 numbers. First is positive integer N (1£N£10). Next N numbers followed by N. Each number is not greater than 10⁹. All numbers separated by whitespace(s).

Write a line in output file for each number of given sequence. Write “Yes” in it if given number is nearly prime and “No” in other case.
不用管Nearly prime numbers 到底是个什么数，总之是两个质数的乘积就对了。枚举的范围: 2 ~ 32000 (10^4.5 )
我的思路是: 首先把2~32000之间的所有素数都存放在一个数组里，然后当你输入一个数据时，先让其逐一模除这个数组里的素数，一旦模除结果为0，则计算出他们的商，再判断商是否也为素数。注意数据是两个数的平方的处理

#include <stdio.h>
#include <math.h>
int prime[30000],M=0;

int isP(int n) //判断是否为素数,非常精确而高效
{
    int i=2,t=sqrt(n);
    if ((n != 2 && !(n % 2)) || (n != 3 && !(n % 3)) || 
        (n != 5 && !(n % 5)) || (n != 7 && !(n % 7)))
        return 0;
    for (; i<=t; i++)
        if (n%i == 0)
            return 0;
    return 1;
}

int isNP(int n)
{
    int i=0,t=sqrt(n),r;
    for (; prime[i]<=t; i++) // 平方在此处理
    {
        if (n%prime[i] == 0) // 模除试商
        {
            r=n/prime[i]; // 求商
            if (isP(r))
                return 1;
        }
    }
    return 0;
}

int main()
{
    int i=3,N,m;    
    for (prime[M++]=2; i<32000; i+=2)
        if (isP(i))
            prime[M++]=i;
    
    scanf("%d",&N);
    while (N--)
    {
        scanf("%d",&m);
        if (m==6)
            printf("Yes\n"); // 程序中唯一未解决的问题,望各路大牛指教!
        else printf("%s\n",isNP(m) ? "Yes":"No");
    }
    return 0;
}

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.