hdu 5055(模拟)

Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1481    Accepted Submission(s): 552

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 a1,a2,a3,⋯,an.(0≤ai≤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)

 

n个数字组成一个数，问能够组成的最大的奇数是多少？不能有前导0

直接模拟：1，如果全是偶数直接输出-1

　　　　   2，从大到小排序，最低位为奇数直接输出，最低位为偶数的话往前挪，找到第一个奇数，注意一下前导0的情况即可。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n,a[];
    char c[];
    while(scanf("%d",&n)!=EOF)
    {
        bool flag = false;
        for(int i=; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]%==) flag = true;
        }
        if(!flag) printf("-1\n");
        else
        {
            int res[];
            sort(a+,a+n+,cmp);
            if(a[n]%==)
            {
                for(int i=; i<=n; i++)
                {
                    res[i] = a[i];
                }
            }
            else
            {
                int t = n;
                for(int i=n; i>=; i--)
                {
                    if(a[i]%==)
                    {
                        t = a[i];
                        a[i] = -;
                        break;
                    }
                }
                int cnt=;
                for(int i=; i<=n; i++)
                {
                    if(a[i]==-) continue;
                    res[cnt++] = a[i];
                }
                res[cnt] = t;
            }
            if(res[]==) printf("-1\n");
            else
            {
                for(int i=; i<=n; i++)
                {
                    printf("%d",res[i]);
                }
                printf("\n");
            }
        }
    }
    return ;
}