A. 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 abc 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 ab and c print the maximum value that you can get.

Input

The input contains three integers ab 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
 

div2的A啊……sb模拟

六种情况搞来搞去

 #include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
LL a,b,c,ans;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int main()
{
a=read();b=read();c=read();
ans=a+b+c;
ans=max(ans,a+b*c);
ans=max(ans,a*b+c);
ans=max(ans,(a+b)*c);
ans=max(ans,a*(b+c));
ans=max(ans,a*b*c);
printf("%lld\n",ans);
}

cf479A

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