主题链接：Expression
1 second
256 megabytes
standard input
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.
The input contains three integers a, b and c,
each on a single line (1 ≤ a, b, c ≤ 10).
Print the maximum value of the expression that you can obtain.
1
2
3
9
2
10
3
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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