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  • GDUFE ACM-1355

    题目:http://acm.gdufe.edu.cn/Problem/read/id/1355

    Problem A. Thickest Burger

    Time Limit: 2000/1000ms (Java/Others)

    Problem Description:

     ACM ICPC is launching a thick burger. The thickness (or the height) of a piece of club steak is A (1 ≤ A ≤ 100). The thickness (or the height) of a piece of chicken steak is B (1 ≤ B ≤ 100). The chef allows to add just three pieces of meat into the burger and he does not allow to add three pieces of same type of meat. As a customer and a foodie, you want to know the maximum total thickness of a burger which you can get from the chef. Here we ignore the thickness of bread, vegetables and other seasonings.

    Input:

    The first line is the number of test cases. For each test case, a line contains two positive integers A and B. 

    Output:

    For each test case, output a line containing the maximum total thickness of a burger. 

    Sample Input:

    10
    68 42
    1 35
    25 70
    59 79
    65 63
    46 6
    28 82
    92 62
    43 96
    37 28
    

    Sample Output:

    178
    71
    165
    217
    193
    98
    192
    246
    235
    102

    思路:判断两个数字哪个大,大的乘以2再加上小的数

    难度:非常简单

    代码:
     1 #include<stdio.h>
     2 int main()
     3 {
     4     int n,a,b,c;
     5     while(scanf("%d",&n)!=EOF)
     6     {
     7         while(n--)
     8         {
     9             scanf("%d %d",&a,&b);
    10             if(a>b)
    11                 c=a*2+b;
    12             else if(a<b)
    13                 c=a+2*b;
    14             else c=a*3;
    15             printf("%d
    ",c);
    16         }
    17     }
    18     return 0;
    19 }
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  • 原文地址:https://www.cnblogs.com/ruo786828164/p/6009302.html
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