HDU 5948 Thickest Burger 【模拟】 (2016ACM/ICPC亚洲区沈阳站)
2016-10-31 15:06
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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)
[align=left]Recommend[/align]
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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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