Problem57

原题：

It is possible to show that the square root of two can be expressed as an infinite continued fraction.
2(开方) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408,
but the eighth expansion, 1393/985,
is the first example where the number of digits
in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions,
how many fractions contain a numerator with more digits than denominator?

 

分析：

根据题意可看出2的开平方的近似序列1为：
 3/2    7/5    17/12   41/29    99/70    239/169     577/408   1393/985 
 可以推导出：
  序列1的递推公式为f(N)=f(N-1)(a-分子,b-分母)=(a+2b)/(a+b)

 

代码如下：

public class Problem57 {//将每组数存进该map中，key-value 对应 分子-分母static List<Scores> scoresList = new ArrayList<Scores>();//将前一千组数放入该数组static{Scores beginNumber = new Scores();beginNumber.setDenominator(BigInteger.valueOf(2));beginNumber.setNumerator(BigInteger.valueOf(3));scoresList.add(beginNumber);while(scoresList.size()<1000){int size = scoresList.size();Scores s = scoresList.get(size-1);scoresList.add(getSores(s));}}//判断分子的位数是否大于分母的位数static boolean exceed(Scores scores){if(String.valueOf(scores.getNumerator()).length()>String.valueOf(scores.getDenominator()).length())return true;return false;}//通过上个分数求下一个分数static Scores getSores(Scores s){Scores next = new Scores();next.setNumerator(s.getDenominator().multiply(BigInteger.valueOf(2)).add(s.getNumerator()));next.setDenominator(s.getDenominator().add(s.getNumerator()));return next;} public static void main(String[] args) {//System.out.println(scoresList);int count = 0;for(Scores s :scoresList){if(exceed(s))count++;}System.out.println("count:"+count);}}class Scores{private BigInteger numerator;//分子private BigInteger denominator;//分母public BigInteger getDenominator() {return denominator;}public void setDenominator(BigInteger denominator) {this.denominator = denominator;}public BigInteger getNumerator() {return numerator;}public void setNumerator(BigInteger numerator) {this.numerator = numerator;}@Overridepublic String toString(){return "该分数的分子为："+this.getNumerator()+" 分母为："+this.getDenominator();} @Overridepublic int hashCode() {final int PRIME = 31;int result = 1;result = PRIME * result + ((denominator == null) ? 0 : denominator.hashCode());result = PRIME * result + ((numerator == null) ? 0 : numerator.hashCode());return result;}@Overridepublic boolean equals(Object obj) {if (this == obj)return true;if (obj == null)return false;if (getClass() != obj.getClass())return false;final Scores other = (Scores) obj;if (denominator == null) {if (other.denominator != null)return false;} else if (!denominator.equals(other.denominator))return false;if (numerator == null) {if (other.numerator != null)return false;} else if (!numerator.equals(other.numerator))return false;return true;}}