几个笔试题目（2010-09-18）

99直线解：

   p(n) = 2         (n=1)
   p(n) = p(n-1)+n  (n>1)

结果好像是：fn(x=999) = 499501

那道钱转换为中文的，贴一个我的解法：用最简方法获得中文，然后再进一步处理中文（“零”问题）

class NumberToChinse {
	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);
		Long number = scanner.nextLong();
		//开始处理
		System.out.println(trim(conversion(number)));
	}

	/*转换*/
	public static String conversion(Long number){
		
		String [] lib1 = {"零","一","二","三","四","五","六","七","八","九"};
		String [] lib2 = {"","十","百","千","万","十","百","千","亿","十","百","千","兆","十","百","千"};
		Map<Integer,String> map1 = new HashMap<Integer,String>();  //汉字映射
		Map<Integer,String> map2 = new HashMap<Integer,String>();  //单位映射
		
		for (int i = 0; i < lib2.length; i++) {
			map2.put(i+1, lib2[i]);
			if (i<10)map1.put(i, lib1[i]);
		}
		char[] str_number = String.valueOf(number).toCharArray();  //得到输入数字的字符串数组
		String result = "";
		for (int i = 0; i < str_number.length; i++) {
			result += map1.get(Integer.valueOf(String.valueOf(str_number[i]))) + map2.get(str_number.length-i);
		}
		return result;
	}
	
	/*处理“零”问题*/
	private static String trim(String result) {
		result = result.replaceAll("(零.零$)+", "");
		result = result.replaceAll("(零.零)+(.)", "零$2");  //可斟酌
		return result;
	}
}

==========

第一道题换种思路
称不让它清零
把1和2放上去，然后把2分别换3，4……10，如果那个重量变化了，就是哪箱，如果都没有变化，就是第一箱
不知道这样行不行

是一种蛮好的思路，不走传统路线。不过在零问题出来上有点问题。比如：1230013转换会成为：一百二十三万零百一十三。

你这样相当于称了很多次，题目要求只能称一次。

String chineseMappingArray = "零...九";
这样不是更好么？

public class MoneyTrans {
	
	private static String[] ChinaDigit = {"零","一","二","三","四","五","六","七","八","九"};
	private static String[] UNIT = {"","","十","百","千"};
	private static String[] BIGUNIT = {"","万","亿"};
	
	private char[] digit;
	public String trans(int n){
		StringBuffer buff = new StringBuffer();
		digit = String.valueOf(n).toCharArray();
		int length = digit.length;		
		int pos = (length - 1)/4;
		int headLength = (length - 1)%4 + 1;		
		buff.append(partTrans(0,headLength) + BIGUNIT[pos--]);
		for (int i = headLength;i < length ; i = i + 4) {
			
			buff.append(partTrans(i , i + 4) +  BIGUNIT[pos--]) ;
		}
		return buff.toString();
	}
	
	private  String partTrans(int start, int end) {
		StringBuffer buff = new StringBuffer();
		boolean isPreDigitZero = false;
		for (int i = start; i < end; i++) {
			int cur = digit[i] - '0';
			if(cur != 0 ){
				if(isPreDigitZero == true){
					buff.append(ChinaDigit[0]);
				}
				buff.append(ChinaDigit[cur] + UNIT[end - i]);
				isPreDigitZero = false;
			}
			else {			
				isPreDigitZero = true;
			}			
		}		
		return buff.toString();		
	}

	public static void main(String[] args) {		
		MoneyTrans transtor = new MoneyTrans();
		String money = transtor.trans(23005602);
		System.out.println(money);
	}

}

这不是高斯定理吗，高斯8岁的时候就证明过了吧！。。。

高斯有这样的定理？我孤弱寡闻了。一般这么证明：（1）（2）a1=2成立，假设ak = a(k-1) +k成立，证明a(k+1) = ak + (k+1)成立。这里的问题主要是，如何证明这步？也就是如何确定知道当从k条直线变为k+1条直线增加了多少部分。