hdu 5055 Bob and math problem

最新推荐文章于 2024-07-20 14:45:00 发布

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#C++ #计数排序 #基本算法

本文介绍了一道算法题目，要求使用给定的数字构造出满足特定条件的最大奇数。通过计数排序找到可用的奇数，并按值降序排列剩余数字来构造答案。

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Bob and math problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1147 Accepted Submission(s): 430

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

Source

BestCoder Round #11 (Div. 2)

题目大意: 求取给出数字组成的最大的奇数
题目分析:计数排序,找最小的奇数,然后从大到小将值顺次填充,特判一下只有一个奇数,剩下的全是0的情况就能过

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;

int n;
int num[11];
int a;

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
      memset ( num , 0 , sizeof ( num ) );
      int end = -1;
      for ( int i = 0 ; i < n ; i++ )
      {
          scanf ( "%d" , &a );
          num[a]++;
      }
      for ( int i = 0 ; i <= 9 ; i++ )
        if ( (i&1)&& num[i] )
        {
            num[i]--;
            end = i;
            break;
        }
      if ( num[0] )
      {
          bool flag = true;
          for ( int i = 1 ; i <= 9 ; i++ )
              if ( num[i] ) flag = false;
          if ( flag ) end = -1;
      }
      if ( end == -1 ) 
      {
          puts ( "-1" );
          continue;
      }
      for ( int i = 9 ; i >= 0 ; i-- )
        while ( num[i] )
        {
            printf ( "%d" , i );
            num[i]--;
        }
      printf ( "%d\n" , end );
    }
}