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

2016-10-31 15:06 531 查看

Thickest Burger

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 63 Accepted Submission(s): 60


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
The first line is the number of test cases. For each test case, a line contains two positive integers A and B.

[align=left]Output[/align]
For each test case, output a line containing the maximum total thickness of a burger.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

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.

[align=left]Source[/align]
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。

1 //
2 //by coolxxx
3 //#include<bits/stdc++.h>
4 #include<iostream>
5 #include<algorithm>
6 #include<string>
7 #include<iomanip>
8 #include<map>
9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-8)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 24
34 #define M 1004
35 using namespace std;
36 typedef long long LL;
37 double anss;
38 LL aans;
39 int cas,cass;
40 int n,m,lll,ans;
41
42 int main()
43 {
44     #ifndef ONLINE_JUDGE
45 //    freopen("1.txt","r",stdin);
46 //    freopen("2.txt","w",stdout);
47     #endif
48     int i,j,k;
49     int x,y,z;
50 //    init();
51     for(scanf("%d",&cass);cass;cass--)
52 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
53 //    while(~scanf("%s",s))
54 //    while(~scanf("%d%d",&n,&m))
55     {
56         scanf("%d%d",&n,&m);
57         printf("%d\n",n+m+max(n,m));
58     }
59     return 0;
60 }
61 /*
62 //
63
64 //
65 */


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