周赛-Expression 分类： 比赛 2015-08-02 09:35 3人阅读 评论(0) 收藏

A. Expression
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 a, b, c 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 a, b and c print the maximum value that you can get.

Input
The input contains three integers a, b 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
枚举飘过

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define LL long long
#define RR freopen("output.txt","r",stdoin)
#define WW freopen("input.txt","w",stdout)

using namespace std;

const int MAX = 100100;

int main()
{
    int a,b,c;
    int Max;
    while(~scanf("%d %d %d",&a,&b,&c))
    {
        Max=0;
        Max=max(Max,a+b+c);
        Max=max(Max,a*(b+c));
        Max=max(Max,(a+b)*c);
        Max=max(Max,a*b*c);
        Max=max(Max,a*b+c);
        Max=max(Max,a+b*c);
        printf("%d
",Max);
    }
    return 0;
}

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