hdu 5463 Clarke and minecraft

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=5463

Clarke and minecraft

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 366    Accepted Submission(s): 193

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a game player of minecraft. 
On that day, Clarke set up local network and chose create mode for sharing his achievements with others. Unfortunately, a naughty kid came his game. He placed a few creepers in Clarke's castle! When Clarke returned his castle without create mode, creepers suddenly blew(what a amazing scene!). Then Clarke's castle in ruins, the materials scattered over the ground. 
Clark had no choice but to pick up these ruins, ready to rebuild. After Clarke built some chests(boxes), He had to pick up the material and stored them in the chests. Clarke clearly remembered the type and number of each item(each item was made of only one type material) . Now Clarke want to know how many times he have to transport at least. 
Note: Materials which has same type can be stacked, a grid can store 64 materials of same type at most. Different types of materials can be transported together. Clarke's bag has 4*9=36 grids.

 

 

Input

The first line contains a number  T(1≤T≤10), the number of test cases. 
For each test case: 
The first line contains a number n, the number of items. 
Then n lines follow, each line contains two integer a,b(1≤a,b≤500), a denotes the type of material of this item, b denotes the number of this material.

 

 

Output

For each testcase, print a number, the number of times that Clarke need to transport at least.

 

 

Sample Input

2

3

2 33

3 33

2 33

10

5 467

6 378

7 309

8 499

5 320

3 480

2 444

8 391

5 333

100 499

 

 

Sample Output

1

2

 

Hint: The first sample, we need to use 2 grids to store the materials of type 2 and 1 grid to store the materials of type 3. So we only need to transport once;

 

 

Source

BestCoder Round #56 (div.2)

 

題目大意：

想要和盒子搬運一些物品，每一個物品都有材質和數量。，一個盒子有36個格子，每一個格子只能裝同一種材質的物品，一個格子最多隻能裝64件物品。最後輸出的是須要搬運幾回。

 

詳見代碼。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int t;
10     int sum[510];
11     scanf("%d",&t);
12     while (t--)
13     {
14         int n,Max=0,ans=0;
15         memset(sum,0,sizeof(sum));
16         scanf("%d",&n);
17         while (n--)
18         {
19             int a,b;
20             scanf("%d%d",&a,&b);
21             sum[a]+=b;//a類有多少個
22             if (a>Max)
23                 Max=a;
24         }
25         for (int i=0; i<=Max; i++)
26         {
27             if (sum[i]==0)
28                 continue;
29             if (sum[i]%64==0)
30                 ans+=sum[i]/64;
31             else
32                 ans+=(sum[i]/64)+1;
33         }
34         int aans;
35         if (ans%36==0)
36             aans=ans/36;
37         else
38             aans=(ans/36)+1;
39         printf ("%d\n",aans);
40     }
41     return 0;
42 }