The 70th problem，UVa10396 Vampire Numbers

今天看Thinking in Java看到一个吸血鬼数的问题，于是查找UVa里也有类似的问题就动手写了先是用Java写的，不过WA了两次，然后没有发现错误，又用c++写的还是不行。最后发现要排序去重。然后改用Java的SortedSet解决了这个问题，主要就是暴力枚举求解。但是同样的算法Java用了将近5秒。c++只用了1秒。差距啊。还有避免不必要的重复的循环不能为00，还有不能为奇数。尽量提高程序的效率。

题目描述：

Problem D

Vampire Numbers

Input: standard input

Output: standard output

Time Limit: 5 seconds

A number v = xy with an even number (n) of digits formed by multiplying a pair of n/2-digit numbers (where the digits are taken from the original number in any order) x and y together is known as vampire number. Pairs of trailing zeros (Both the numbers have a trailing zero) are not allowed. If v is a vampire number then x and y are called its "fangs." Examples of 4-digit vampire numbers include

1) 21 x 60 = 1260
2) 15 x 93 = 1395
3) 35 x 41 = 1435
4) 30 x 51 = 1530
5) 21 x 87 = 1827
6) 27 x 81 = 2187
7) 80 x 86 = 6880

In this program you will have to find all the 4, 6 and
8 digit even vampire numbers.

Input

The input file contains maximum ten lines of input. Each line contains a single integer n whose value is
4, 6 or 8. Input is terminated by end of file.

Output

For each input n produce all the n-digit vampire numbers that are even in ascending order. Print a blank line after the output for each set of input.

Sample Input:

4

4

Sample Output:

1260

1530

6880

1260

1530

6880

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

int main()
{
    bool isNum(int a,int b);
    int n;
    int range[] = {1,10,100,1000,10000};
    int set[10][10000],flag[10];
    memset(set,0,sizeof(range));
    memset(flag,0,sizeof(flag));
    while(scanf("%d",&n) != -1)
    {
        if(flag[n] == -1)
       {
        int k = 0;
        for(int i=range[n / 2 - 1]; i<range[n / 2]; i++)
        {
            for(int j=i; j<range[n / 2]; j++)
            {
                if(i % 2 != 0 && j % 2 != 0)continue;
                if(i % 10 == 0 && j % 10 == 0)continue;
                if(isNum(i,j))
                {
                    set[n][k] = i * j;
                    k++;
                }
            }
        }

       }
       flag[n] = -1;
       for(int i=0; ; i++)
       {
          if(set[n][i] == 0)break;
          printf("%d",set[n][i]);
       }
       printf("\n");

    }

}
bool isNum(int a,int b)
{
    int arr[10];
    int mult = a * b;
    while(a != 0)
    {
        arr[a % 10]++;
			a /= 10;
    }
    while(b != 0)
    {
        arr[b % 10]++;
			b /= 10;
    }
    while(mult != 0)
    {
        arr[mult % 10]--;
			mult /= 10;
    }
   for(int i=0; i<10; i++){
			if(arr[i] != 0)
				return false;
		}
		return true;
}
Java代码：

<pre class="java" name="code">import java.util.*;

public class Main10396 {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int[] range = {1,10,100,1000,10000};
		Object[] temp = new Object[100000];
		int[] flag = new int[10];
		SortedSet<Integer> set4 = new TreeSet<Integer>();
		SortedSet<Integer> set6 = new TreeSet<Integer>();
		SortedSet<Integer> set8 = new TreeSet<Integer>();
		while(scan.hasNext()){
			int n = scan.nextInt();
			int k = 0;
			if(flag[n] != -1){
			for(int i=range[n / 2 - 1]; i<range[n / 2]; i++){
				for(int j=i; j<range[n / 2]; j++){
					if((i % 2 != 0)&& (j % 2 != 0))continue;//除去奇数
					if(i % 10 == 0 && j % 10 == 0)continue;//除去00
					if(isNum(i,j) == true){
						if(n == 4){
							set4.add(i * j);
						}
						if(n == 6){
							set6.add(i * j);
						}
						if(n == 8){
							set8.add(i * j);
						}
					}
				}
			}
			}
			flag[n] = -1;
			if(n == 4){
				for(int i : set4){
					System.out.println(i);
				}
			}
			if(n == 6){
				for(int i : set6){
					System.out.println(i);
				}
			}
			if(n == 8){
				for(int i : set8){
					System.out.println(i);
				}
			}
			System.out.println();

		}

	}
	public static boolean isNum(int a,int b){
		int[] arr = new int[10];
		int mult = a * b;
		while(a != 0){
			arr[a % 10]++;
			a /= 10;
		}
		while(b != 0){
			arr[b % 10]++;
			b /= 10;
		}
		while(mult != 0){
			arr[mult % 10]--;
			mult /= 10;
		}
		for(int i=0; i<10; i++){
			if(arr[i] != 0)
				return false;
		}
		return true;
	}

}

最后说一句Tinking in Java 讲的不错。