主题链接:Expression

Expression
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers abc on
the blackboard. The task was to insert signs of operations '+' and '*',
and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:

  • 1+2*3=7
  • 1*(2+3)=5
  • 1*2*3=6
  • (1+2)*3=9

Note that you can insert operation signs only between a and b,
and between b and c, that is, you cannot swap integers.
For instance, in the given sample you cannot get expression (1+3)*2.

It's easy to see that the maximum value that you can obtain is 9.

Your task is: given ab and c print
the maximum value that you can get.

Input

The input contains three integers ab and c,
each on a single line (1 ≤ a, b, c ≤ 10).

Output

Print the maximum value of the expression that you can obtain.

Sample test(s)
input
1
2
3
output
9
input
2
10
3
output
60

大致题意:a, b, c三个数。在三个数中,插入“+” 和“*”运算符的随意两个组合,求能组成的表达式的值得最大值。(能够用括号)

解题思路:没啥说的。直接暴力,总共就6种组合。

AC代码:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff int x[9]; int main()
{
// #ifdef sxk
// freopen("in.txt","r",stdin);
// #endif
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
x[0] = a + b + c;
x[1] = a + (b * c);
x[2] = a * (b + c);
x[3] = (a + b) * c;
x[4] = (a * b) + c;
x[5] = a * b * c;
sort(x, x+6);
printf("%d\n",x[5]);
}
return 0;
}

版权声明:本文sxk原创文章。转载本文,请添加链接^_^

Codeforces Round #274 (Div. 2) --A Expression的更多相关文章

  1. codeforces水题100道 第八题 Codeforces Round #274 (Div. 2) A. Expression (math)

    题目链接:http://www.codeforces.com/problemset/problem/479/A题意:给你三个数a,b,c,使用+,*,()使得表达式的值最大.C++代码: #inclu ...

  2. Codeforces Round #274 (Div. 2) 解题报告

    题目地址:http://codeforces.com/contest/479 这次自己又仅仅能做出4道题来. A题:Expression 水题. 枚举六种情况求最大值就可以. 代码例如以下: #inc ...

  3. Codeforces Round #274 (Div. 2)

    A http://codeforces.com/contest/479/problem/A 枚举情况 #include<cstdio> #include<algorithm> ...

  4. Codeforces Round #274 (Div. 1) C. Riding in a Lift 前缀和优化dp

    C. Riding in a Lift Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/pr ...

  5. Codeforces Round #274 (Div. 1) B. Long Jumps 数学

    B. Long Jumps Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/ ...

  6. Codeforces Round #274 (Div. 1) A. Exams 贪心

    A. Exams Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/480/problem/A Des ...

  7. Codeforces Round #274 (Div. 2)-C. Exams

    http://codeforces.com/contest/479/problem/C C. Exams time limit per test 1 second memory limit per t ...

  8. Codeforces Round #274 Div.1 C Riding in a Lift --DP

    题意:给定n个楼层,初始在a层,b层不可停留,每次选一个楼层x,当|x-now| < |x-b| 且 x != now 时可达(now表示当前位置),此时记录下x到序列中,走k步,最后问有多少种 ...

  9. Codeforces Round #274 (Div. 2) E. Riding in a Lift(DP)

    Imagine that you are in a building that has exactly n floors. You can move between the floors in a l ...

随机推荐

  1. 本地或者服务器同时启动2个或多个tomcat

    一,修改配置文件server.xml的端口 C:\apache-tomcat-5.5.23-1\conf\server.xml用记事本什么的打开修改3个地方   第一: <Server port ...

  2. Cordova CLI源码分析(二)——package.json

    每个包需要在其顶层目录下包含一个package.json文件,该文件不仅是包的说明,也影响npm安装包时的配置选项 更多参数详见参考文档https://npmjs.org/doc/json.html ...

  3. 怎样让你的安卓手机瞬间变Firefox os 畅玩firefox os 应用

    Firefox os 手机迟迟不能在国内大面积上市.如今能买到的Firefox os手机国内就一款Firefox os ZET OPEN C ,但这款手机配置确实还不如人意.价格方面也不实惠,对于我们 ...

  4. Memcached在.net中的应用

    一.MemCached下载 服务端下载:http://memcachedproviders.codeplex.com/ client下载:path=/trunk">http://sou ...

  5. VMware vSphere 服务器虚拟化之二十六 桌面虚拟化之View Persona Management

    VMware vSphere 服务器虚拟化之二十六 桌面虚拟化之View Persona Management 实验失败告终,启动VMware View Persona Management服务报10 ...

  6. Windows phone 8 学习笔记(2) 数据文件操作

    原文:Windows phone 8 学习笔记(2) 数据文件操作 Windows phone 8 应用用于数据文件存储访问的位置仅仅限于安装文件夹.本地文件夹(独立存储空间).媒体库和SD卡四个地方 ...

  7. python学习笔记之五:抽象

    本文会介绍如何将语句组织成函数,还会详细介绍参数和作用域的概念,以及递归的概念及其在程序中的用途. 一. 创建函数 函数是可以调用,它执行某种行为并且返回一个值.用def语句即可定义一个函数:(并非所 ...

  8. MySQL Windows ZIP 免费安装和启动设置

    MySQL Windows ZIP免安装版,设置和启动的过程事实上挺麻烦的.以下一步一步介绍使用的过程: 1.下载Windows (x86, 64-bit), ZIP Archive: 2.解压zip ...

  9. NETSH WINSOCK RESET这个命令的意义和效果?

    简要地netsh winsock reset命令含义复位 Winsock 文件夹.一机多用的假设Winsock协议配置问题,那么问题会导致网络连接,我们需要使用netsh winsock reset命 ...

  10. Windows 2008 配置ASP+ACCESS环境(亲身体会)

    我们公司OA系统是用asp开发的,时间有些长了,原来只是公司总部,部署到内网就可以了,现在要求全国各地的分公司也要用,而且接入了56短网的短信接口(http://www.56dxw.com),主要起到 ...