题目链接

Problem Description

Ignatius drinks milk everyday, now he is in the supermarket and he wants to choose a bottle of milk. There are many kinds of milk in the supermarket, so Ignatius wants to know which kind of milk is the cheapest.

Here are some rules:

\1. Ignatius will never drink the milk which is produced 6 days ago or earlier. That means if the milk is produced 2005-1-1, Ignatius will never drink this bottle after 2005-1-6(inclusive).

\2. Ignatius drinks 200mL milk everyday.

\3. If the milk left in the bottle is less than 200mL, Ignatius will throw it away.

\4. All the milk in the supermarket is just produced today.

Note that Ignatius only wants to buy one bottle of milk, so if the volumn of a bottle is smaller than 200mL, you should ignore it.

Given some information of milk, your task is to tell Ignatius which milk is the cheapest.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.

Each test case starts with a single integer N(1<=N<=100) which is the number of kinds of milk. Then N lines follow, each line contains a string S(the length will at most 100 characters) which indicate the brand of milk, then two integers for the brand: P(Yuan) which is the price of a bottle, V(mL) which is the volume of a bottle.

Output

For each test case, you should output the brand of the milk which is the cheapest. If there are more than one cheapest brand, you should output the one which has the largest volume.

Sample Input

2
2
Yili 10 500
Mengniu 20 1000
4
Yili 10 500
Mengniu 20 1000
Guangming 1 199
Yanpai 40 10000

Sample Output

Mengniu
Mengniu HintIn the first case, milk Yili can be drunk for 2 days, it costs 10 Yuan. Milk Mengniu can be drunk for 5 days, it costs 20 Yuan. So Mengniu is the cheapest.In the second case,
milk Guangming should be ignored. Milk Yanpai can be drunk for 5 days, but it costs 40 Yuan. So Mengniu is the cheapest.

分析:

简单的结构体排序,要找到最简单的牛奶,肯定就是单价最少的。

代码:

#include<iostream>
#include<stdio.h>
#include<cstdlib>
#include<string>
#include<string.h>
#include<algorithm>
using namespace std;
struct Niunai
{
char str[103];
double a;///价钱
double b;///体积
double c;///单价
} aa[104];
bool camp(Niunai x,Niunai y)
{
if(x.c!=y.c)///单价按照从小到大排序
return (x.c<y.c);
else
return (x.b>y.b);///单价相同,按照体积从大到小排序
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
memset(aa,0,sizeof(aa));///每次进行完,数组要刷新,
int m;
double n1;
scanf("%d",&m);
for(int i=0; i<m; i++)
{
scanf("%s%lf%lf",aa[i].str,&aa[i].a,&aa[i].b);
if(aa[i].b<200)
aa[i].c=0x3f3f3f;
else
{
n1=(int)(aa[i].b/200);
if(n1>=5)
n1=5;
aa[i].c=(double)(aa[i].a/n1);///得到每天的价钱
}
}
sort(aa,aa+m,camp);
printf("%s\n",aa[0].str);
}
return 0;
}

HDU 1070 Milk (模拟)的更多相关文章

  1. hdu 1070 Milk(贪心)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1070 Milk Time Limit: 2000/1000 MS (Java/Others)    M ...

  2. HDU 1070 - Milk

    给每种牛奶价格和量 要求买最便宜的牛奶 #include <iostream> using namespace std; int t,n; ][]; ],v[]; int main() { ...

  3. HDU 4121 Xiangqi 模拟题

    Xiangqi Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4121 ...

  4. hdu 5071 Chat(模拟)

    题目链接:hdu 5071 Chat 题目大意:模拟题. .. 注意最后说bye的时候仅仅要和讲过话的妹子说再见. 解题思路:用一个map记录每一个等级的妹子讲过多少话以及是否有这个等级的妹子.数组A ...

  5. hdu 4740【模拟+深搜】.cpp

    题意: 给出老虎的起始点.方向和驴的起始点.方向.. 规定老虎和驴都不会走自己走过的方格,并且当没路走的时候,驴会右转,老虎会左转.. 当转了一次还没路走就会停下来.. 问他们有没有可能在某一格相遇. ...

  6. HDU 2568[前进]模拟

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2568 关键思想:傻傻地模拟 代码如下: #include<iostream> using ...

  7. hdu 4964 恶心模拟

    http://acm.hdu.edu.cn/showproblem.php?pid=4964 给定语句,按照语法翻译html并输出. 就是恶心的模拟,递归搞就行了 处理id和class时,在一个'&g ...

  8. HDOJ.1070 Milk(贪心)

    Milk 点我挑战题目 题意分析 每组测试数据给出一系列牛奶商品,分别是牛奶的品牌,价格,以及体积.在读取数据的时候,体积在200以下的牛奶直接忽略掉.并且每天要喝200ML的牛奶.但是无论牛奶体积有 ...

  9. HDU 5965 枚举模拟 + dp(?)

    ccpc合肥站的重现...一看就觉得是dp 然后强行搞出来一个转移方程 即 根据第i-1列的需求和i-1 i-2列的枚举摆放 可以得出i列摆放的种类..加了n多if语句...最后感觉怎么都能过了..然 ...

随机推荐

  1. Swift-闭包理解(二)

    简明扼要的闭包表达式 其实Swift已经为我们提供了很多简化的语法,可以让我们保证代码的高可读性和维护性.还用上面的例子来说明,对于  greetPeople 这个全局函数来说,其实只需要使用一次,所 ...

  2. PokeCats开发者日志(十三)

      现在是PokeCats游戏开发的第六十二天的晚上,把软著权登记证书的截图加上,又重新提交审核了一遍,但愿能过吧...

  3. 多个表单数据提交下的serialize()应用

    在实际开发场景中,难免遇到需要多个表单的数据传递问题. 之所以要进行多表单的数据传递是因为可以进行数据分组,便于数据的维护. 这个时候,出于不依赖jquery的考虑,有一个原生js函数来解决这个问题无 ...

  4. win7 安装 MongoDB 及简单操作

    下载地址 http://dl.mongodb.org/dl/win32/x86_64 这里用的版本是 mongodb-latest-signed.msi 同时下载 mongodb-compass 下载 ...

  5. 【Windows】Windows Restart Manager 重启管理器

    Restart Manager(以下简称RM)可以减少或避免安装或更新程序所需要的系统重启次数.安装(或更新)过程中需要重启的主要原因是需要更新的某些文件当前正被一些其它程序或服务所使用.RM允许除关 ...

  6. BZOJ4299 Codechef FRBSUM(主席树)

    感觉非常不可做,于是考虑有什么奇怪的性质. 先考虑怎么求子集和mex.将数从小到大排序,假设已经凑出了0~n的所有数,如果下一个数>n+1显然mex就是n+1了,否则若其为x则可以凑出1~n+x ...

  7. 在Windows*上编译Tensorflow教程

    背景介绍 最简单的 Tensorflow 的安装方法是在 pip 一键式安装官方预编译好的包 pip install tensorflow 通常这种预编译的包的编译参数选择是为了最大兼容性而不是为了最 ...

  8. 【BZOJ5296】【CQOI2018】破解D-H协议(BSGS)

    [BZOJ5296][CQOI2018]破解D-H协议(BSGS) 题面 BZOJ 洛谷 Description Diffie-Hellman密钥交换协议是一种简单有效的密钥交换方法.它可以让通讯双方 ...

  9. React Render Callback Pattern(渲染回调模式)

    React Render Callback Pattern,渲染回调模式,其实是将this.props.children当做函数来调用. 例如: 要根据user参数确定渲染Loading还是Profi ...

  10. JavaScript in 操作符

    JavaScript的in操作符可以用来判断一个属性是否属于一个对象,也可以用来变量一个对象的属性 1. 判断属性属于对象 var mycar = {make: "Honda", ...