hdu5463(基础)

Clarke and minecraft

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 864    Accepted Submission(s): 430

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

//hdu5463(基础)
//题目大意:有n件物品,(可能有种类相同的物品)现在有一个揹包，其内部有36个单元,每个单元可以放同类型
//的物品64个,问将所有物品从一处运到另一处需要多少次？
//解题思路:定义一个数组,将同类的物品归到一起.先求出每种物品需要多少个单元的总和,然后因为运一次也就是
//36个单元,所以除以36就是需要运的次数,不过需要注意物品不足放满36个单元,也需要一次. 
#include<stdio.h>
#include<string.h>
int main()
{
   int a[510],t,n,i,u,v,k;
   scanf("%d",&t);
   while(t--)
   {
   	  memset(a,0,sizeof(a));
      scanf("%d",&n);	
      for(i=0;i<n;i++)
      {
      	scanf("%d%d",&u,&v);
      	a[u]+=v;
	  }
	  k=0;
	  for(i=1;i<=500;i++)
	  {
	  	if(a[i]>0)
	  	{
	  	  if(a[i]%64>0) k+=a[i]/64+1;
		  else          k+=a[i]/64;
		}
	  }
	  if(k%36>0) k=k/36+1;
	  else       k=k/36;
	  printf("%d\n",k);
   }
   return 0;	
}