Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14398 Accepted Submission(s): 3551
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.
Author
Ignatius.L
題目大意就是讓給你牛奶的品牌，花費的錢，體積。讓你判斷買哪個最合算。注意幾點：體積小於兩百的不喝，每天喝的是兩百ML，牛奶最多喝五千。下面上代碼。附帶詳細的解析。
#include
#include
struct node //結構體裡面的信息
{
char name[105]; // 這是牛奶品牌的名字
int v; // 牛奶的體積
int money;// 花費的錢
}
f[105]; // 用來存放以上三個變量
void solve()
{
int n,m,i,j,p,max; //定義變量 p表示喝的天數，n代表有幾個案例,m代表m個品牌牛奶的名字、價格、體積
while(scanf("%d",&n)!=EOF)
{
while(n--)
{
float sum;// 注意：這裡必須用folat 或者double因為（17/5 19/5的值是一樣的，所以必須用浮點數，來保證每天喝的價格是不同）
float min=999999999;// 讓它等於一個很大的數，以便來找最小數。
sum=0;j=0;max=-1;// 讓max等於一個很小的數，以便來找最大數。
scanf("%d",&m);// 輸入
for(i=1;i<=m;i++)
scanf("%s%d%d",f[i].name,&f[i].money,&f[i].v);// 輸入牛奶品牌，花的錢，體積、
for(i=1;i<=m;i++)
{
if(f[i].v<200)// 如果小於兩百則繼續（小於兩百就不喝了）
continue;
if(f[i].v>1000)// 如果大於一千 則讓P=5，因為最多喝五天。
p=5;//
else
p=f[i].v/200;// 否則除以兩百就是喝的天數
sum=float(f[i].money/p);// 注意，我就是因為這裡WA了許多次。（不能寫成float(f[i].money）/p；這樣得到的數是整數！！）
if(sumf[j].v)// 則尋找體積大的奶。
{
j=i;//把變量賦值給j，就是體積大的品牌名
}
}
}
printf("%s\n",f[j].name);// 輸出
}
}
}
int main()
{
solve();// 解決問題。
return 0;
}
