HDU 5948 Thickest Burger 【模拟】 (2016ACM/ICPC亚洲区沈阳站)

Thickest Burger

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 63    Accepted Submission(s): 60

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

1068 421 3525 7059 7965 6346 628 8292 6243 9637 28

 

Sample Output

1787116521719398192246235102
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.
 

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5948

题目大意：

　　给a,b求a+b+max(a,b)

题目思路：

　　【模拟】

　　水题，直接模拟即可。

////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#pragma comment(linker,"/STACK:1024000000,1024000000")#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-8)#define J 10000#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#define N 24#define M 1004using namespace std;typedef long long LL;double anss;LL aans;int cas,cass;int n,m,lll,ans;int main(){#ifndef ONLINE_JUDGE//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;int x,y,z;//init();for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))//while(~scanf("%d%d",&n,&m)){scanf("%d%d",&n,&m);printf("%d\n",n+m+max(n,m));}return 0;}/*////*/