hdu 5055（坑）

题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5055

Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 695    Accepted Submission(s): 263

Problem Description

Recently, Bob has been thinking about a math problem.

There are N Digits, each digit is between 0 and 9. You need to use this N Digits to constitute an Integer.

This Integer needs to satisfy the following conditions:

• 1. must be an odd Integer.
• 2. there is no leading zero.
• 3. find the biggest one which is satisfied 1, 2.

Example:

There are three Digits: 0, 1, 3. It can constitute six number of Integers. Only "301", "103" is legal, while "130", "310", "013", "031" is illegal. The biggest one of odd Integer is "301".

 

Input

There are multiple test cases. Please process till EOF.

Each case starts with a line containing an integer N ( 1 <= N <= 100 ).

The second line contains N Digits which indicate the digit $a_1, a_2, a_3, \cdots, a_n. ( 0 \leq a_i \leq 9)$.

 

Output

The output of each test case of a line. If you can constitute an Integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.

 

Sample Input

3
0 1 3
3
5 4 2
3
2 4 6

 

Sample Output

301
425
-1

 

Source

field=problem&key=BestCoder+Round+%2311+%28Div.+2%29&source=1&searchmode=source">BestCoder Round #11 (Div. 2)

 

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思路:这题有点略坑~思路挺简单。可是细心才干AC，

直接将n个数排序。然后找最小的奇数移出就可以。

PS：（1）要注意n==1的情况

（2）You need to use this N Digits to constitute an Integer.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cmath>
const int INF=99999999;
#include <algorithm>
using namespace std;

int a[110];
bool cmp(int a,int b)
{
  return a>b;
}
int main()
{
    int n;
    while(cin>>n)
    {
      for(int i=1;i<=n;i++)
      {
        cin>>a[i];
      }
     if(n==1)
     {
       if(a[1]&1)
            cout<<a[1]<<endl;
       else
            cout<<-1<<endl;
       continue;
     }
     sort(a+1,a+1+n,cmp);

     int flag=INF;
     for(int i=n;i>=1;i--)
     {
       if(a[i]&1)
       {
        flag=i;
        break;
       }
     }
     if(flag==INF)
     {
       cout<<-1<<endl;
       continue;
     }
     if(flag==1&&a[2]==0)
     {
       cout<<-1<<endl;
       continue;
     }
     for(int i=1;i<=n;i++)
     {
       if(i!=flag)
            cout<<a[i];
     }
     cout<<a[flag]<<endl;
    }
    return 0;
}