hdu 5463 Clarke and minecraft
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
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.
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.
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;
#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
- hdu 5463 Clarke and minecraft(贪心)
Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke turned i ...
- 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 ...
- 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 ...
- 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? ...
- 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 ...
- 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? ...
- 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 ...
- 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 ...
- 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
- springmvc+json 前后台数据交互
1. 配置(1) 文件配置参考这里(2) 导入jackson相关包:jackson-annotations-2.9.4.jar,jackson-core-2.9.4.jar,jackson-datab ...
- Avito Cool Challenge 2018 自闭记
A:n==2?2:1. #include<iostream> #include<cstdio> #include<cmath> #include<cstdli ...
- P3455 [POI2007]ZAP-Queries
题目描述 Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He ha ...
- Day24-KindEditor基本使用和文件操作1
KindEditor是一套开源的HTML可视化编辑器,主要用于让用户在网站上获得所见即所得编辑效果,兼容IE.Firefox.Chrome.Safari.Opera等主流浏览器. 摘自老师博客:htt ...
- MVC如何设置启动页
1.解决方案下的项目,右键,属性,Web,特定页,切换下其他选项以保存
- Android热修复原理(一)热修复框架对比和代码修复
在Android应用开发中,热修复技术被越来越多的开发者所使用,也出现了很多热修复框架,比如:AndFix.Tinker.Dexposed和Nuwa等等.如果只是会这些热修复框架的使用那意义并不大,我 ...
- 【linux之文件查看,操作,权限管理】
一.shell如何处理命令 1.shell会根据在命令中出现的空格字符,将命令划分为多个部分 2.判断第一个字段是内部命令还是外部命令 内部命令:内置于shell的命令(shell builtin) ...
- 常用Build-in Keywords
1. Variables |- Set variable |- Create list |- Evaluate |- Get Variable Value 2. Conditional |- Run ...
- C/C++ 移位计算代替乘除运算
测试移位和乘除的比较,发现移位比乘除运算快一个位数的速度,但是难点在于判断是否是2的幂次级的数,如果不是还得通过代码拆分到2的幂次+上分子的累和,然后通过移位得到2的次幂数这样; 下列代码只是简单的判 ...
- 科学计算三维可视化---TVTK管线与数据加载(用IVTK根据观察管线)
一:用IVTK根据观察管线 (一)引入该工具 from tvtk.tools import ivtk 可能需要安装pygments pip3 install pygments (二)使用ivtk显示立 ...