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最新推荐文章于 2020-05-15 11:04:37 发布

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本文介绍了一种算法，该算法旨在从一组数字中构造出最大的整数，该整数能够同时被2、3和5整除。文章详细解释了算法的工作原理，包括如何选择数字以确保最终结果符合除法条件，特别是关于如何通过删除某些数字来解决模3余数的问题。

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yxc loves math lessons very much, so he doesn't attend them, unlike wwl. But now yxc wants to get a good mark for math. For that Ms. lxh, his ACM teacher, gave him a new task. yxc solved the task immediately. Can you?

You are given a set of digits, your task is to find the maximum integer that you can make from these digits. The made number must be divisible by 2, 3, 5 without a residue. It is permitted to use not all digits from the set, it is forbidden to use leading zeroes.

Each digit is allowed to occur in the number the same number of times it occurs in the set.

Input
A single line contains a single integer n (1 ≤ n ≤ 100000) — the number of digits in the set. The second line contains n digits, the digits are separated by a single space.

Output
On a single line print the answer to the problem. If such number does not exist, then you should print -1.

Example
Input
1
0
Output
0
Input
11
3 4 5 4 5 3 5 3 4 4 0
Output
5554443330
Input
8
3 2 5 1 5 2 2 3
Output
-1
Note (You Will Win)

In the first sample there is only one number you can make — 0. In the second sample the sought number is 5554443330. In the third sample it is impossible to make the required number.

题意差不多是最多用 n个数字组成一个最大的数，这个数可以被2 5 3 整除。以为着末尾数字必须为0，并且各位之和为3的倍数。

题目卡在了可以删除一个数，也可以删除两个数上。

如果全部的和取模3 余1 那么删除一个取膜余1 的或者两个取模余2 的都可以解决问题。

如果取模3 余2 那么可以删除一个取模余2 或者 2个取模余1 的。

如果无法满足上述，倘若有0 输出0 没0的话输出-1

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main ()
{
    int n,i,j;
    int  a[15];
    memset(a,0,sizeof(a));
    cin >>n;
    int sum=0;
    for(i=0; i<n; i++)
    {
        cin >>j;
        a[j]++;
        sum+=j;
    }
    if(a[0]==0)
    {
        printf("-1\n");
    }
    else
    {
       int f=0;
       if(sum%3==0) f=1;
       else if(sum%3==1)
       {
         for(i=1;i<=9;i=i+3)
         {
            if(a[i]>=1)
             {
                a[i]--;
                 f=1;
                break;
             }
         }
         if(f==0)
         {
            for(i=2;i<=9;i=i+3)
            {
                if(a[i]>=1)
                {
                    a[i]--;
                    if(a[i]>=1)
                    {
                        a[i]--;
                        f=1;
                        break;
                    }
                    else if(a[i+3]>=1)
                    {
                        a[i+3]--;
                         f=1;
                        break;
                    }
                    else if(a[i+6]>=1)
                    {
                        a[i+6]--;
                         f=1;
                        break;
                    }
                    else
                        f=0;
                }
            }
         }
       }
        else if(sum%3==2)
       {
         for(i=2;i<=9;i=i+3)
         {
            if(a[i]>=1)
             {
                a[i]--;
                 f=1;
                break;
             }
         }
         if(f==0)
         {
            for(i=1;i<=9;i=i+3)
            {
                if(a[i]>=1)
                {
                    a[i]--;
                    if(a[i]>=1)
                    {
                        a[i]--;
                        f=1;
                        break;
                    }
                    else if(a[i+3]>=1)
                    {
                        a[i+3]--;
                         f=1;
                        break;
                    }
                    else if(a[i+6]>=1)
                    {
                        a[i+6]--;
                         f=1;
                        break;
                    }
                    else
                        f=0;
                }
            }
         }
       }
        if(f)
        {
            int z=1;
            int y=0;
            for(i=9; i>=0; i--)
            {
                while (a[i]--)
                {
                    if(i!=0)
                    {
                        printf("%d",i);
                        y=1;
                        z=0;
                    }
                    else
                    {
                        if(y==1)
                            printf("%d",i);
                    }

                }
            }
            if(z)
                printf("0\n");
        }

    }
    return 0;


}