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  • Codeforces Round #274 (Div. 2)-A. Expression

    http://codeforces.com/contest/479/problem/A

    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

    解题思路:求出a,b,c之间加上+或者*号之后表达式的最大值,可以加上括号,所以直接暴力

     1 #include<stdio.h>
     2 
     3 int main()
     4 {
     5 
     6     int a,b,c, sum;
     7     int max;
     8     while(scanf("%d%d%d",&a,&b, & c) != EOF){
     9         max = -1;
    10         sum = a + b * c;
    11         max = max > sum ? max : sum;
    12         sum = (a + b) * c;
    13         max = max > sum ? max : sum;
    14         sum = a * b + c;
    15         max = max > sum ? max : sum;
    16         sum = a * (b +c);
    17         max = max > sum ? max : sum;
    18         sum = a + b + c;
    19         max = max > sum ? max : sum;
    20         sum = a * b * c;
    21         max = max > sum ? max : sum;
    22         printf("%d ", max);
    23     }
    24     return 0;

    25 } 

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  • 原文地址:https://www.cnblogs.com/angle-qqs/p/4044698.html
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