• HDU 5055 Bob and math problem(结构体)


    主题链接:http://acm.hdu.edu.cn/showproblem.php?pid=5055



    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


    官方题解:



    代码例如以下:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int main()
    {
        int n;
        int a[117];
        while(~scanf("%d",&n))
        {
            int minn = 10;
            int flag = 0;
            int tt = 0;
            for(int i = 0; i < n; i++)
            {
                scanf("%d",&a[i]);
                if(a[i]&1)
                {
                    flag = 1;
                    if(a[i] < minn)
                    {
                        minn = a[i];
                        tt = i;
                    }
                }
            }
            a[tt] = 10;
            sort(a,a+n);
            if(!flag)
            {
                printf("-1
    ");
                continue;
            }
            flag = 0;
            for(int i = n-2; i >= 0; i--)
            {
                if(a[i]==0 && !flag)
                    continue;
                flag = 1;
                printf("%d",a[i]);
            }
            if(flag || n==1)
                printf("%d
    ",minn);
            else
                printf("-1
    ");
        }
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/mengfanrong/p/4840521.html
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