题目链接: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
 
题目大意:
想要和盒子搬运一些物品,每个物品都有材质和数量。,一个盒子有36个格子,每个格子只能装同一种材质的物品,一个格子最多只能装64件物品。最后输出的是需要搬运几次。
 
详见代码。
 #include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int main()
{
int t;
int sum[];
scanf("%d",&t);
while (t--)
{
int n,Max=,ans=;
memset(sum,,sizeof(sum));
scanf("%d",&n);
while (n--)
{
int a,b;
scanf("%d%d",&a,&b);
sum[a]+=b;//a类有多少个
if (a>Max)
Max=a;
}
for (int i=; i<=Max; i++)
{
if (sum[i]==)
continue;
if (sum[i]%==)
ans+=sum[i]/;
else
ans+=(sum[i]/)+;
}
int aans;
if (ans%==)
aans=ans/;
else
aans=(ans/)+;
printf ("%d\n",aans);
}
return ;
}

hdu 5463 Clarke and minecraft的更多相关文章

  1. hdu 5463 Clarke and minecraft(贪心)

    Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke turned i ...

  2. HDU 5628 Clarke and math——卷积,dp,组合

    HDU 5628 Clarke and math 本文属于一个总结了一堆做法的玩意...... 题目 简单的一个式子:给定$n,k,f(i)$,求 然后数据范围不重要,重要的是如何优化这个做法. 这个 ...

  3. BestCoder Round #56/hdu5463 Clarke and minecraft 水题

    Clarke and minecraft 问题描述 克拉克是一名人格分裂患者.某一天,克拉克分裂成了一个游戏玩家,玩起了minecraft.渐渐地,克拉克建起了一座城堡. 有一天,克拉克为了让更多的人 ...

  4. hdu 5565 Clarke and baton 二分

    Clarke and baton Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php? ...

  5. hdu 5563 Clarke and five-pointed star 水题

    Clarke and five-pointed star Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/show ...

  6. hdu 5465 Clarke and puzzle 二维线段树

    Clarke and puzzle Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php? ...

  7. hdu 5464 Clarke and problem dp

    Clarke and problem Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php ...

  8. HDU 5628 Clarke and math dp+数学

    Clarke and math 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5628 Description Clarke is a patient ...

  9. HDU 5464 Clarke and problem 动态规划

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5464 Clarke and problem  Accepts: 130  Submissions: ...

随机推荐

  1. 删除log日志中包含某个字符的行

    sed -i '/{Str}/d' abc.txt 假如你的log日志中某行有sleep字符,直接输入命令: sed -i '/sleep/d' log.log 如果删除的是一个变量的值,假如是var ...

  2. java的finally用法

    finally作为异常处理的一部分,它只能用在try/catch语句中,并且附带一个语句块,表示这段语句最终一定会被执行(不管有没有抛出异常),经常被用在需要释放资源的情况下. 之前在写爬虫的时候数据 ...

  3. ThinkPHP 5.x远程命令执行漏洞分析与复现

    0x00 前言 ThinkPHP官方2018年12月9日发布重要的安全更新,修复了一个严重的远程代码执行漏洞.该更新主要涉及一个安全更新,由于框架对控制器名没有进行足够的检测会导致在没有开启强制路由的 ...

  4. android.database.CursorIndexOutOfBoundsException:Index -1 requested, with a size of 1(zz)

    android.database.CursorIndexOutOfBoundsException:Index -1 requested, with a size of 1 http://blog.cs ...

  5. Java类编译、加载、和执行机制

    Java类编译.加载.和执行机制 标签: java 类加载 类编译 类执行 机制 0.前言 个人认为,对于JVM的理解,主要是两大方面内容: Java类的编译.加载和执行. JVM的内存管理和垃圾回收 ...

  6. NOIWC前的交流题目汇总

    RT 2018.12.27 i207M:BZOJ 4695 最假女选手 以维护最大值为例,记录最大值和严格次大值和最大值的出现次数,然后取min的时候递归到小于最大值但大于次大值修改,这个就是最重要的 ...

  7. OS X 安装pyspider

    pyspider安装的过程中,需要安装pycurl.有几个坑 一.首先遇到权限的问题 因为/Library目录是root权限,所以非root用户对该目录的读写经常会遇到权限问题,但是不宜切换成root ...

  8. 4:JAVA UUID 生成

    GUID是一个128位长的数字,一般用16进制表示.算法的核心思想是结合机器的网卡.当地时间.一个随即数来生成GUID.从理论上讲,如果一台机器每秒产生10000000个GUID,则可以保证(概率意义 ...

  9. 详细MATLAB 中BP神经网络算法的实现

    MATLAB 中BP神经网络算法的实现 BP神经网络算法提供了一种普遍并且实用的方法从样例中学习值为实数.离散值或者向量的函数,这里就简单介绍一下如何用MATLAB编程实现该算法. 具体步骤   这里 ...

  10. Linux可执行文件后缀问题

    一般来说,可执行文件没有扩展名. Linux不根据扩展名判断文件类型,而是根据文件的内容来判断.所以扩展名的作用是帮助人来识别文件,对于Linux系统本身来说没有什么用处. .sh结尾表示是shell ...