HDU 5055 Bob and math problem

Bob and math problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 961 Accepted Submission(s): 368


[align=left]Problem Description[/align]
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".

[align=left]Input[/align]
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 [Math Processing Error].

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

301
425
-1

[align=left]Source[/align]
BestCoder Round #11 (Div. 2)

算法分析：给你n个数，让着n个数组成 符合要求的数字输出，组不出来的话就输出-1.
条件1. 组成的是 奇数。
条件2. 没有前导0。
条件3.在满足前两个条件的基础上保证数最大！

算法实现：1. 检查这n个数里 有没有奇数， 如果没有 直接输出-1 。
2. 检查0的个数， 如果0的个数>=n-1，这样组出来的数 要么具有前导0，要么不是奇数，直接输出-1.
3. 将最小的奇数 放在最后输出， 将其余的数字 按从大到小的顺序输出即可。
4. 如果n==1，也就是说，只输入一个数，奇数的话就直接输出，否则输出-1。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>

using namespace std;

int main()
{
int n, mm, nn;
int a[101];
int ff[101];
int i, j;
int ee, oo, dd;

while(scanf("%d", &n)!=EOF)
{
if(n==1)
{
scanf("%d", &nn);
if(nn%2==0)
{
printf("-1\n");
continue;
}
else if(nn%2==1 )
{
printf("%d\n", nn);
continue;
}

}

ee=0; oo=0; dd=0;
mm=999999;
for(i=0; i<n; i++)
{
scanf("%d", &a[i] );
if(a[i]%2==1)
{
oo++; //奇数
if( a[i]<mm )
{
mm=a[i];
}
}
else if(a[i]%2==0)
{
ee++;
if(a[i]==0)
dd++; //0的个数
}
}
if(oo==0)
{
printf("-1\n");
continue;
}
if(dd>=n-1)
{
printf("-1\n");
continue;
}
memset(ff, 0, sizeof(ff));
sort(a, a+n);
for(i=0; i<n; i++)
{
if(a[i]==mm)
{
ff[i]=1;
break;
}
}
for(i=n-1; i>=0; i--)
{
if(ff[i]==0)
printf("%d", a[i] );
}
printf("%d\n", mm);
}
return 0;
}