Bob and math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 401    Accepted Submission(s): 149

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
 
贪心
代码:
 #include<cstdio>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iostream>
using namespace std;
int str[];
int main()
{
int n,i,res;
while(scanf("%d",&n)!=EOF)
{
res=;
for(i=;i<n;i++)
{
scanf("%d",&str[i]);
if(str[i]&==&&res>str[i])
res=str[i];
}
if(res==){
printf("-1\n");
continue;
}
sort(str,str+n,greater<int>());
int fir=-,st=res;
for(i=;i<n;i++)
{
if(res==str[i]) res=-;
else
{
fir=str[i];
break;
}
}
if(fir==) printf("-1\n");
else
{
if(fir!=-)
printf("%d",fir);
for( i++; i<n;i++)
if(res==str[i])res=-;
else
printf("%d",str[i]);
printf("%d\n",st);
}
}
return ;
}

hdu----(5055)Bob and math problem(贪心)的更多相关文章

  1. 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 think ...

  2. 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 ...

  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

    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 ...

  5. HDU 4974 A simple water problem(贪心)

    HDU 4974 A simple water problem pid=4974" target="_blank" style="">题目链接 ...

  6. 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 ...

  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. eclipse选中变量,相同变量高亮。

    选择Windows->Preferences->Java->Editor->Mark Occurrences,全部选择并保存. 如下图:

  2. How to: Update an .edmx File when the Database Changes

    https://msdn.microsoft.com/en-us/library/cc716697.aspx In the Model Browser, right-click the .edmx f ...

  3. Spring的"Hello, world",还有"拿来主义"

    这里有两个类: com.practice包下的SpringTest.java和PersonService.java. Spring可以管理任意的POJO(这是啥?),并不要求Java类是一个标准的Ja ...

  4. hdu 5023 A Corrupt Mayor's Performance Art 线段树

    A Corrupt Mayor's Performance Art Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100 ...

  5. Nginx入门笔记之————配置文件结构

    在nginx.conf的注释符号位# nginx文件的结构,这个对刚入门的同学,可以多看两眼. 默认的config: #user nobody; worker_processes ; #error_l ...

  6. SAP接口编程 之 JCo3.0系列(04) : 会话管理

    在SAP接口编程之 NCo3.0系列(06) : 会话管理 这篇文章中,对会话管理的相关知识点已经说得很详细了,请参考.现在用JCo3.0来实现. 1. JCoContext 如果SAP中多个函数需要 ...

  7. rpm and yum commands

    rpm命令 rpm包,由“-”.“.”构成,包名.版本信息.版本号.运行平台 对已安装软件信息的查询 rpm -qa                             查询已安装的软件 rpm ...

  8. 关于php的一些小知识

    浏览目录: 一.PHP的背景和优势: 二.PHP原理简介: 三.PHP运行环境配置: 四.编写简单的PHP代码以及测试. 一.PHP的背景和优势 1.1   什么是PHP? PHP是能让你生成动态网页 ...

  9. (五)uboot移植补基础之shell

    1.shell介绍:shell是操作系统的终端命令行 (1)shell可以理解为软件系统提供给用户操作的命令行界面,可以说它是人机交互的一种方式.(2)我们可以使用shell和操作系统.uboot等软 ...

  10. Java 14 类型信息

    14 类型信息 运行是识别对象和类的信息 两种方式RTTI 假定编译时已经知道所有的类型反射 运行时发现和使用类的信息 1 RTTI //多态 创建一个具体的对象(Circle Square Tria ...