codeforces Round #274(div2) A解题报告

最新推荐文章于 2022-11-22 23:15:14 发布

西湖醋瓜

最新推荐文章于 2022-11-22 23:15:14 发布

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本文介绍了一个简单的算法问题，即如何通过合理安排运算符使由三个正整数构成的算术表达式的值达到最大。文章提供了具体的例子并给出了实现这一目标的代码。

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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

题目大意：

给三个数，中间可以加*或者是+，使得数字最大，不能改变顺序，但可以加括号。

解法：

根据题意，我们可以得知就6种不同的组合：

a+b+c;

a*b*c;

a+b*c;

(a+b)*c;

a*b+c;

a*(b+c);

然后找最大值就可以了。

也可以简化一下： a, b, c均为正整数

a+b*c = (a/c+b)*c, a/c <= a -> a+b*c <= (a+b)*c;

a*b+c = a*(b+c/a) c/a <= c -> a*b+c <= a*(b+c); 这样就只有4个式子了

代码：

#include <cstdio>
#include <algorithm>

using namespace std;

int a, b, c;
int x[10];

int main() {
	scanf("%d%d%d", &a, &b, &c);

	x[1] = a+b+c;
	x[2] = a*b*c;
	x[3] = (a+b)*c;
	x[4] = a*(b+c);

	sort(x+1, x+7);

	printf("%d\n", x[6]);
}