題目：http://poj.org/problem?id=3197

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分析：對于分數利用歐幾里德算法可以寫成連分數的形式，這樣p和q很大，用Java大數寫很方便。

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import java.io.*; import java.util.*; import java.math.BigInteger;public class Main {static final int N = 100005;static BigInteger ans[] = new BigInteger[N];static int cnt = 0;static boolean EqualToZero(BigInteger x){if(x.compareTo(BigInteger.ZERO) == 0) return true;else return false;}static void gcd(BigInteger a,BigInteger b){cnt = 0;ans[cnt++] = a.divide(b);BigInteger r = a.mod(b);while(!EqualToZero(r)){a = b;b = r;ans[cnt++] = a.divide(b);r = a.mod(b);}ans[cnt++] = BigInteger.ONE;ans[cnt-2] = ans[cnt-2].subtract(BigInteger.ONE);}public static void main(String[] args){Scanner cin = new Scanner(System.in);int tt = 1;while(cin.hasNextBigInteger()){BigInteger p = cin.nextBigInteger();BigInteger q = cin.nextBigInteger();if(EqualToZero(p) && EqualToZero(q)) break;gcd(p,q);System.out.println("Case "+(tt++)+":");System.out.println(p+" / "+q);int length = 3*(cnt-1);for(int i=0;i<cnt;i++)length += ans[i].toString().length();int len = length;int ct = 0;for(int k=0;k<cnt-1;k++){int pos = ans[k].toString().length() + 3;int record = pos;int num = length - pos;int tmp = 0;if(num % 2 == 1){tmp = num / 2 + 1;pos += tmp;}else{tmp = num / 2;pos += tmp;}for(int i=1;i<=ct;i++)System.out.print(".");for(int i=1;i<=length;i++){if(i == pos) System.out.print("1");else System.out.print(".");}System.out.println("");for(int i=1;i<=ct;i++)System.out.print(".");System.out.print(ans[k]+".+.");for(int i=1;i<=num;i++)System.out.print("-");System.out.println("");length -= record;ct += record;}for(int i=1;i<len;i++)System.out.print(".");System.out.println("1");}} }

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