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hdu 5055（坑）

2024-07-25 03:34:22   219人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5055

Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 695    Accepted Submission(s): 263


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
BestCoder Round #11 (Div. 2)
 

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思路:这题有点略坑~思路挺简单，但是细心才能AC，

直接将n个数排序，然后找最小的奇数移出即可；

PS：（1）要注意n==1的情况

          （2）You need to use this N Digits to constitute an Integer.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cmath>
const int INF=99999999;
#include <algorithm>
using namespace std;

int a[110];
bool cmp(int a,int b)
{
  return a>b;
}
int main()
{
    int n;
    while(cin>>n)
    {
      for(int i=1;i<=n;i++)
      {
        cin>>a[i];
      }
     if(n==1)
     {
       if(a[1]&1)
            cout<<a[1]<<endl;
       else
            cout<<-1<<endl;
       continue;
     }
     sort(a+1,a+1+n,cmp);

     int flag=INF;
     for(int i=n;i>=1;i--)
     {
       if(a[i]&1)
       {
        flag=i;
        break;
       }
     }
     if(flag==INF)
     {
       cout<<-1<<endl;
       continue;
     }
     if(flag==1&&a[2]==0)
     {
       cout<<-1<<endl;
       continue;
     }
     for(int i=1;i<=n;i++)
     {
       if(i!=flag)
            cout<<a[i];
     }
     cout<<a[flag]<<endl;
    }
    return 0;
}


hdu 5055（坑）


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