[5055]Bob and math problem（hdu）

Bob and math problem

Problem Description
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

30 1 335 4 232 4 6

 

Sample Output

301425-1

 

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>int main(){    int a[110];    int n,i,j,k,m,t;    int flag;    while(~scanf("%d",&n))    {        flag=0;        for(i=0; i<n; i++)            scanf("%d",&a[i]);        for(i=0; i<n-1; i++)            for(j=0; j<n-i-1; j++)                if(a[j]<a[j+1])                {                    t=a[j];                    a[j]=a[j+1];                    a[j+1]=t;                }        for(i=n-1; i>=0; i--)        {            if(a[i]%2!=0)            {                k=i;                m=a[i];                flag=1;                break;            }        }        if(flag==0)            puts("-1");        else        {            for(i=k; i<n-1; i++)            {                a[i]=a[i+1];            }            a[n-1]=m;            if(a[0]==0)//前导是0则输出“-1”            {                puts("-1");            }                        else            {                for(i=0; i<n; i++)                printf("%d",a[i]);                printf("\n");            }                                }    }    return 0;}