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

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

全栈程序员-站长 • 2022年1月11日下午1:00 • 未分类 • 阅读 47

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主题链接：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

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;
}

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