hdu 5055（坑）

題目連結：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 BestCoder Round #11 (Div. 2)  

Recommend heyang   |   We have carefully selected several similar problems for you:   5057  5056  5054  5053  5052  思路:這題有點略坑~思路挺簡單，但是細心才能AC，

直接将n個數排序，然後找最小的奇數移出即可；

PS：（1）要注意n==1的情況

          （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;
}