2016ACM/ICPC亚洲区沈阳站-HDU-5948-Thickest Burger（最水的签到题）

Thickest Burger

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 50 Accepted Submission(s): 48

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 breads, 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
Hint
Consider the first test case, since 68+68+42 is bigger than 68+42+42 the answer should be 68+68+42 = 178.
Similarly since 1+35+35 is bigger than 1+1+35, the answer of the second test case should be 1+35+35 = 71.

开场一小时四道题的秘密

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>#include<queue>using namespace std;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int a,b;        scanf("%d%d",&a,&b);        printf("%d\n",max(a,b)*2+min(a,b));    }    return 0;}