Codeforces Round #274 (Div. 2) A. Expression

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

题意：求操作的最大值

思路：水的搜索

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main() {
	int a, b, c;
	scanf("%d%d%d", &a, &b, &c);
	int Max = -1;
	int s1 = a + b + c;
	Max = max(Max, s1);
	int s2 = a + b * c;
	Max = max(Max, s2);
	int s3 = a * b + c;
	Max = max(Max, s3);
	int s4 = a * b * c;
	Max = max(Max, s4);
	int s5 = (a + b) * c;
	Max = max(Max, s5);
	int s6 = a * (b + c);
	Max = max(Max, s6);
	printf("%d\n", Max);	
	return 0;
}