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主题链接：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
代码语言：javascript复制1
2
3output
代码语言：javascript复制9input
代码语言：javascript复制2
10
3output
代码语言：javascript复制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：
代码语言：javascript复制#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("%dn",x[5]);
}
return 0;
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