Codeforces Round #274 (Div. 2) A Expression_html/css_WEB-ITnose

题目链接：expression

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Expression

time limit per test

memory limit per test

input

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:

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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

123

output

input

2103

output

60

大致题意：a, b, c三个数，在三个数中，插入“+” 和“*”运算符的任意两个组合，求能组成的表达式的值得最大值。（可以用括号）

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 0x7fffffffint 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;}