# Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 401    Accepted Submission(s): 149

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

 1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #include<functional>
5 #include<iostream>
6 using namespace std;
7 int str[120];
8 int main()
9 {
10   int n,i,res;
11   while(scanf("%d",&n)!=EOF)
12   {
13       res=10;
14       for(i=0;i<n;i++)
15     {
16        scanf("%d",&str[i]);
17       if(str[i]&1==1&&res>str[i])
18           res=str[i];
19     }
20     if(res==10){
21         printf("-1\n");
22         continue;
23     }
24     sort(str,str+n,greater<int>());
25      int fir=-1,st=res;
26     for(i=0;i<n;i++)
27     {
28         if(res==str[i]) res=-1;
29         else
30         {
31             fir=str[i];
32             break;
33         }
34     }
35     if(fir==0) printf("-1\n");
36     else
37     {
38         if(fir!=-1)
39           printf("%d",fir);
40         for( i++; i<n;i++)
41         if(res==str[i])res=-1;
42         else
43          printf("%d",str[i]);
44         printf("%d\n",st);
45     }
46   }
47   return 0;
48 }

0 条评论

## 相关文章

### poj----Ubiquitous Religions

Ubiquitous Religions Time Limit: 5000MS Memory Limit: 65536K Total Submi...

3676

2277

### poj---（2886）Who Gets the Most Candies?（线段树+数论）

Who Gets the Most Candies? Time Limit: 5000MS Memory Limit: 131072K Tota...

2865

### HDOJ 1003

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J...

22010

### HDUOJ 1099——Lottery

Lottery Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J...

3647

### HDUOJ--1159Common Subsequence

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536...

2697

### POJ----The Suspects

The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: ...

36412

2925

3985

### hdu 4009 Transfer water（最小型树图）

Transfer water Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/657...

3689