POJ3197(连分数表示)
题目：http://poj.org/problem?id=3197
分析：对于分数利用欧几里德算法可以写成连分数的形式，这样p和q很大，用Java大数写很方便。
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"); } }}