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
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