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：

#include<stdio.h>
int main()
{
    int n,a,b,c;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            scanf("%d %d",&a,&b);
            if(a>b)
                c=a*2+b;
            else if(a<b)
                c=a+2*b;
            else c=a*3;
            printf("%d\n",c);
        }
    }
    return 0;
}