cf479A Expression

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 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

 

 

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=0,f=1;char ch=getchar();  
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}  
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';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);  
}