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2013
11-10

POJ 2196 Specialized Four-Digit Numbers [解题报告] Java

Specialized Four-Digit Numbers

问题描述 :

Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.

For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 1*1728 + 8*144 + 9*12 + 3, its duodecimal representation is 189312, and these digits also sum up to 21. But in hexadecimal 2991 is BAF16, and 11+10+15 = 36, so 2991 should be rejected by your program.

The next number (2992), however, has digits that sum to 22 in all three representations (including BB016), so 2992 should be on the listed output. (We don’t want decimal numbers with fewer than four digits — excluding leading zeroes — so that 2992 is the first correct answer.)

输入:

There is no input for this problem

输出:

Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.

样例输入:

There is no input for this problem

样例输出:

2992
2993
2994
2995
2996
2997
2998
2999
...

解题代码:

//* @author popop0p0popo
public class Main{
	public static void main(String[] args){
		int t,h,w;
		for (int i=2992;i< 10000 ;i++ ){
			t=getI(i);
			h=getD(Integer.toHexString(i));
			w=getD(Integer.toString(i,12));
			if (t==h&&h==w){
				System.out.println(i);
			}
		}
	}

	public static int getD(String n){
		int d=0;
		for (int i=0;i< n.length() ;i++ ){
			if (Character.isLetter(n.charAt(i))){
				d=d+10+(int)(n.charAt(i)-'a');
			}
			else{
				d=d+(int)(n.charAt(i)-'0');
			}
		}
		return d;
	}

	public static int getI(int n){
		int a=n/1000;
		int b=(n-1000*a)/100;
		int c=(n-1000*a-100*b)/10;
		int d=n-1000*a-100*b-10*c;
		return a+b+c+d;
	}
}

  1. 你的理解应该是:即使主持人拿走一个箱子对结果没有影响。这样想,主持人拿走的箱子只是没有影响到你初始选择的那个箱子中有奖品的概率,但是改变了其余两个箱子的概率分布。由 1/3,1/3 变成了 0, 2/3