【CODEFORCES】 A. Expression

A. 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 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

题解：DIV2的A题...算出所有情况，取最大值即可。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int a,b,c,ans;

int main()
{
    scanf("%d%d%d",&a,&b,&c);
    ans=a+b+c;
    if (ans<a+b*c) ans=a+b*c;
    if (ans<(a+b)*c) ans=(a+b)*c;
    if (ans<a*b+c) ans=a*b+c;
    if (ans<a*(b+c)) ans=a*(b+c);
    if (ans<a*b*c) ans=a*b*c;
    printf("%d",ans);
    return 0;
}