hdu 5055 Bob and math problem（模拟）

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本文介绍了一种算法问题，即如何从给定的一组0到9的数字中构造出满足特定条件的最大奇数整数。具体条件包括：该整数必须为奇数、不能有前导零，并且是在所有可能组合中数值最大的。文章通过示例说明了问题，并提供了一段C++代码实现，展示了如何筛选和排序数字以获得所需结果。

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

Sample Output

301
425
-1

题意：给你n个0~9的数字，问你这些数字组合成的数中满足1：为奇数 2：没有前导零的最大数

思路：直接模拟即可。最后一位取最小的奇数，第一位取最大的非零数，中间的数就从大到小取了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 110
int a[N],b[N];
bool cmp(int c,int d)
{
    return c>d;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        int odd=111,v;
        for(int i=0; i<n; i++)
            if(a[i]&1&&a[i]<odd)
            {
                odd=a[i];
                v=i;
            }
        if(odd==111) printf("-1\n");
        else
        {
            b[n-1]=odd;
            a[v]=-1;
            int maxn=-1;
            for(int i=0; i<n; i++)
                if(a[i]>0&&a[i]>maxn)
                {
                    maxn=a[i];
                    v=i;
                }
            if(maxn==-1&&n>1)
            {
                printf("-1\n");
                continue;
            }
            if(n>1)
            {
                b[0]=maxn;
                a[v]=-1;
                sort(a,a+n,cmp);
                int cnt=1;
                for(int i=0; i<n; i++)
                {
                    if(a[i]==-1) continue;
                    b[cnt++]=a[i];
                }
            }
            for(int i=0;i<n;i++)
                printf("%d",b[i]);
            printf("\n");
        }
    }
    return 0;
}