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:

watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvdTAxMjg2MDA2Mw==/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast" alt="">

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\n");
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\n",minn);
else
printf("-1\n");
}
return 0;
}

版权声明:本文博主原创文章。博客,未经同意不得转载。

HDU 5055 Bob and math problem(结构体)的更多相关文章

  1. HDU 5055 Bob and math problem(简单贪心)

    http://acm.hdu.edu.cn/showproblem.php?pid=5055 题目大意: 给你N位数,每位数是0~9之间.你把这N位数构成一个整数. 要求: 1.必须是奇数 2.整数的 ...

  2. hdu 5055 Bob and math problem

    先把各个数字又大到小排列,如果没有前导零并且为奇数,则直接输出.如果有前导零,则输出-1.此外,如果尾数为偶数,则从后向前找到第一个奇数,并把其后面的数一次向前移动,并把该奇数放到尾部. 值得注意的是 ...

  3. hdu 5055 Bob and math problem (很简单贪心)

    给N个数字(0-9),让你组成一个数. 要求:1.这个数是奇数 2.这个数没有前导0 问这个数最大是多少. 思路&解法: N个数字从大到小排序,将最小的奇数与最后一位交换,把剩下前N-1位从大 ...

  4. hdu----(5055)Bob and math problem(贪心)

    Bob and math problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Oth ...

  5. HDU 1757 A Simple Math Problem 【矩阵经典7 构造矩阵递推式】

    任意门:http://acm.hdu.edu.cn/showproblem.php?pid=1757 A Simple Math Problem Time Limit: 3000/1000 MS (J ...

  6. hdu 1757 A Simple Math Problem (矩阵快速幂)

    Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 ...

  7. hdu 1757 A Simple Math Problem (乘法矩阵)

    A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  8. HDU 5615 Jam's math problem

    Jam's math problem Problem Description Jam has a math problem. He just learned factorization.He is t ...

  9. HDU 1757 A Simple Math Problem (矩阵快速幂)

    题目 A Simple Math Problem 解析 矩阵快速幂模板题 构造矩阵 \[\begin{bmatrix}a_0&a_1&a_2&a_3&a_4&a ...

随机推荐

  1. jquery ui 笔记

    准备: 1.下载jquery ui库:http://jqueryui.com/download/ 2.选择theme 3.建立一个良好的发展环境(mysql.netbeans) 4.创建数据库:jqu ...

  2. 开源 java CMS - FreeCMS2.1公布

    项目地址:http://www.freeteam.cn/ FreeCMS商业版V2.1更新功能 1.web页面信息採集:通过简单配置就可以抓取目标网页信息,支持增量式採集.keyword替换.定时採集 ...

  3. Action、Category、Data、Extras知识具体解释

    开头 Intent作为联系各Activity之间的纽带,其作用并不仅仅仅仅限于简单的数据传递.通过其自带的属性,事实上能够方便的完毕非常多较为复杂的操作.比如直接调用拨号功能.直接自己主动调用合适的程 ...

  4. BC 2015在百度之星程序设计大赛 - 预赛(1)(矩形区域-旋转卡)

    矩形区域 Accepts: 717 Submissions: 1619 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 ...

  5. A星寻路lua实现

    他遇见了自己的主动性的需要找到它的项目的方式,我决定开始学习A明星,为A星我没有深究,它只能说是勉强获得需求.在此和大家分享一下.共同进步. A星有一个公式 f(x) = g(x) + h(x) ,这 ...

  6. jquery之ajax中国乱码的解决方案

    $.ajax({ dataType : 'json',type : 'POST',url : 'http://localhost/test/test.do',data : {id: 1, type: ...

  7. Java TCP/UDP网络通信编程

    本文转自:http://www.cnblogs.com/cdtarena/archive/2013/04/10/3012282.html 网络应用中基本上都是TCP(Transmission Cont ...

  8. Phoenix(sql on hbase)简单介绍

    Phoenix(sql on hbase)简单介绍 介绍: Phoenix is a SQL skin over HBase delivered as a client-embedded JDBC d ...

  9. NET MVC权限验证

    ASP.NET MVC权限验证 封装类 写该权限类主要目地 为了让权限配置更加的灵活,可以根据SQL.json.或者XML的方式来动态进行页面的访问控制,以及没有权限的相关跳转. 使用步骤 1.要建一 ...

  10. flex eclipse综合spring入门

    首先下载FlashBuilder_4_7_LS10_win64.exe试了几eclipse安装没有成功插头,含有myeclipse8.5.spring sts2.9.2.eclipse3.5.j2ee ...