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  • 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 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
    
     

    div2的A啊……sb模拟

    六种情况搞来搞去

     1 #include<cstdio>  
     2 #include<iostream>  
     3 #include<cstring>  
     4 #include<cstdlib>  
     5 #include<algorithm>  
     6 #include<cmath>  
     7 #include<queue>  
     8 #include<deque>  
     9 #include<set>  
    10 #include<map>  
    11 #include<ctime>  
    12 #define LL long long  
    13 #define inf 0x7ffffff  
    14 #define pa pair<int,int>  
    15 #define pi 3.1415926535897932384626433832795028841971  
    16 using namespace std;  
    17 LL a,b,c,ans;  
    18 inline LL read()  
    19 {  
    20     LL x=0,f=1;char ch=getchar();  
    21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}  
    22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}  
    23     return x*f;  
    24 }  
    25 int main()  
    26 {  
    27     a=read();b=read();c=read();  
    28     ans=a+b+c;  
    29     ans=max(ans,a+b*c);  
    30     ans=max(ans,a*b+c);  
    31     ans=max(ans,(a+b)*c);  
    32     ans=max(ans,a*(b+c));  
    33     ans=max(ans,a*b*c);  
    34     printf("%lld
    ",ans);  
    35 }  
    cf479A
    ——by zhber,转载请注明来源
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  • 原文地址:https://www.cnblogs.com/zhber/p/4055120.html
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