(Relax 贪心1.4)POJ 2325 Persistent Numbers(使用贪心策略解决这么一个问题: 给定一个数n，求一个最小的数m,使得m的各位的乘积==n)

import java.math.BigInteger;import java.util.Scanner;public class POJ_2325 {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);BigInteger n;BigInteger zero = new BigInteger("0");BigInteger nOne = new BigInteger("-1");BigInteger ten = new BigInteger("10");int a[] = new int[4000];while(scanner.hasNext()){n = scanner.nextBigInteger();if(n.compareTo(nOne) == 0){break;}if(n.compareTo(ten) < 0){//如果这个数<10System.out.println("1" + n);continue;}boolean flag = false;int count = 0;while(true){//如果这个数>=10int i;flag = false;for(i = 9 ; i >= 2 ; --i){//贪心策略:要使获得的数最小,则从最低位开始枚举最大数if(n.mod(BigInteger.valueOf(i)).compareTo(zero) == 0){flag = true;n = n.divide(BigInteger.valueOf(i));a[++count] = i;break;}}if(n.compareTo(ten) < 0){//如果这个数已经<10,则处理已经结束flag = true;a[++count] = n.intValue();break;}if(flag == false){//如果这个数没有2~9之间的因数&&>10,则分解失败break;}}if(flag == false){System.out.println("There is no such number.");}else{int i;for(i = count ; i >= 1;--i){//逆序输出System.out.print(a[i]);}System.out.println();}}}}