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

Hello everyone , I meet you again , I'm the king of the whole stack .

Topic links ：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 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

Copy

output

9

Copy

input

2
10
3

Copy

output

60

Copy

General meaning ：a, b, c Three numbers . Among the three numbers , Insert “+” and “*” Any two combinations of operators , Find the maximum value of the expression composed of energy .（ Can use brackets ）

Their thinking ： Nothing to say. . Direct violence , All in all 6 Combinations of .

AC Code ：

#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;
}

Copy

Copyright notice ： this paper sxk Original article . Reprint this article , Please add a link ^_^

Publisher ： Full stack programmer stack length , Reprint please indicate the source ：https://javaforall.cn/117058.html Link to the original text ：https://javaforall.cn