杭电OJ 1070 milk
2016-05-27 19:23
330 查看
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.
#include <iostream>
#include <string>
#include <math.h>
using namespace std;
class Min
{
public:
double rate;
string name;
int volumn;
Min()
{
rate=10000000;
name='\0';
volumn=-1;
}
void Clear()
{
rate=10000000;
name='\0';
volumn=-1;
}
};
int main()
{
freopen("D:\\input.txt","r",stdin);
freopen("D:\\output.txt","w",stdout);
int n,m;
string name;
int price,volumn,day;
double rate;
Min min;
Min s_min;
bool flag=false;
cin>>n;
while(n--)
{
min.Clear();
s_min.Clear();
flag=false;
cin>>m;
while(m--)
{
cin>>name;
cin>>price;
cin>>volumn;
if(volumn<200)
{
rate=(double)price/day;
if(rate<s_min.rate||((abs(rate-min.rate)<1e-6)&&volumn>s_min.volumn))
{
s_min.name=name;
s_min.rate=rate;
s_min.volumn=volumn;
}
}
else
{
flag=true;
day=volumn/200;
if(day>5)
day=5;
rate=(double)price/day;
if(min.rate>rate||(abs(min.rate-rate)<1e-6&&volumn>min.volumn))
{
min.rate=rate;
min.name=name;
min.volumn=volumn;
}
}
}
if(flag==true)
cout<<min.name<<endl;
else
cout<<s_min.name<<endl;
}
}
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.<
4000
br />
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.
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.
#include <iostream>
#include <string>
#include <math.h>
using namespace std;
class Min
{
public:
double rate;
string name;
int volumn;
Min()
{
rate=10000000;
name='\0';
volumn=-1;
}
void Clear()
{
rate=10000000;
name='\0';
volumn=-1;
}
};
int main()
{
freopen("D:\\input.txt","r",stdin);
freopen("D:\\output.txt","w",stdout);
int n,m;
string name;
int price,volumn,day;
double rate;
Min min;
Min s_min;
bool flag=false;
cin>>n;
while(n--)
{
min.Clear();
s_min.Clear();
flag=false;
cin>>m;
while(m--)
{
cin>>name;
cin>>price;
cin>>volumn;
if(volumn<200)
{
rate=(double)price/day;
if(rate<s_min.rate||((abs(rate-min.rate)<1e-6)&&volumn>s_min.volumn))
{
s_min.name=name;
s_min.rate=rate;
s_min.volumn=volumn;
}
}
else
{
flag=true;
day=volumn/200;
if(day>5)
day=5;
rate=(double)price/day;
if(min.rate>rate||(abs(min.rate-rate)<1e-6&&volumn>min.volumn))
{
min.rate=rate;
min.name=name;
min.volumn=volumn;
}
}
}
if(flag==true)
cout<<min.name<<endl;
else
cout<<s_min.name<<endl;
}
}
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.<
4000
br />
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.
相关文章推荐
- 简单的四则运算
- 数的奇偶性
- ACMer博客瀑布流分析
- ACM程序设计大赛题目分类
- 计算字符串最后一个单词长度
- ACM网址
- 1272 小希的迷宫
- 1272 小希的迷宫
- hdu 1250 大数相加并用数组储存
- 矩阵的乘法操作
- 蚂蚁爬行问题
- 蚂蚁爬行问题
- 求两个数的最大公约数【ACM基础题】
- 打印出二进制中所有1的位置
- 杭电题目---一只小蜜蜂
- HDOJ 1002 A + B Problem II (Big Numbers Addition)
- 初学ACM - 半数集(Half Set)问题 NOJ 1010 / FOJ 1207
- 初学ACM - 组合数学基础题目PKU 1833
- POJ ACM 1002
- 高精度加法——杭电1002