排列组合在工作中的实际运用

个人理解：这个问题的本质是，每一行不选和选1，对于乘积的结果是没有影响的，所以乘积结果取决于你选了多少个3和多少个9，同时注意到3和9的关系，选了一个9也就相当于选了两个三，也就是最终乘积的不同结果的个数取决于能够选到多少个3，能选到多少个9，也就是有多少个不同的结果，那就是看个最值
设有x个3和y个9，显然是当x=0，y=6的时候可以取到最大，也就是说最多可以选到6*2=12个3，这12个值每个都能取道，所以，最终结果是original score共有12种值

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楼上的说到点子上去了,,但实际上是 14种值,,你忘记加上 1 和 0 了.
我用程序"暴力"破解的结果是这样的:
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

/**
* @author Xujin.Jiao
*
*/
public class Caculate {
    private static final int[] pf = {0, 1, 3, 9};
    private static final int[] cost = {0, 1, 3, 9};
    private static final int[] ofDay = {0, 1, 3, 9};
    private static final int[] impact = {0, 1, 3, 9};
    private static final int[] sr = {0, 1, 9};
    private static final int[] bd = {0, 1, 9};

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int totalSels = 0;
        int value = 0;
        List<Integer> valueList = new ArrayList<Integer>();
       
        totalSels = pf.length * cost.length * ofDay.length *
                    impact.length * sr.length * bd.length;
       
        for (int i = 0; i < pf.length; i++) {
            for (int j = 0; j < cost.length; j++) {
                for (int k = 0; k < ofDay.length; k++) {
                    for (int l = 0; l < impact.length; l++) {
                        for (int m = 0; m < sr.length; m++) {
                            for (int n = 0; n < bd.length; n++) {
                                value = (0 == pf[i] ? 1 : pf[i]) * (0 == cost[j] ? 1 : cost[j]) * (0 == ofDay[k] ? 1 : ofDay[k]) *
                                        (0 == impact[l] ? 1 : impact[l]) * (0 == sr[m] ? 1 : sr[m]) * (0 == bd[n] ? 1 : bd[n]);
                                if (0 == pf[i] && 0 == cost[j] && 0 == ofDay[k] &&
                                    0 == impact[l] && 0 == sr[m] && 0 == bd[n]) {
                                    value = 0;
                                }
                                if (!valueList.contains(new Integer(value))) {
                                    valueList.add(new Integer(value));
                                }
                            }
                        }
                    }
                }
            }
        }

        System.out.println("There are " + totalSels + " different selections.");
        System.out.println("There are " + valueList.size() + " different values.");
       
        System.out.println("These values are listed as follows:");
        Collections.sort(valueList);
        for (int i = 0; i < valueList.size(); i++) {
            System.out.println(valueList.get(i).toString());
        }
    }

}




There are 2304 different selections.
There are 14 different values.
These values are listed as follows:
0
1
3
9
27
81
243
729
2187
6561
19683
59049
177147
531441