A. Expression
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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
枚举飘过
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define LL long long
#define RR freopen("output.txt","r",stdoin)
#define WW freopen("input.txt","w",stdout)
using namespace std;
const int MAX = 100100;
int main()
{
int a,b,c;
int Max;
while(~scanf("%d %d %d",&a,&b,&c))
{
Max=0;
Max=max(Max,a+b+c);
Max=max(Max,a*(b+c));
Max=max(Max,(a+b)*c);
Max=max(Max,a*b*c);
Max=max(Max,a*b+c);
Max=max(Max,a+b*c);
printf("%d\n",Max);
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。