HDU 1070 一道结构体的题目

只能说，我被这道题快折磨疯了，，一点信心都没了。Wrong answer接近两个小时，WA的我好心碎了。不过最后还是AC了。果然还是粗心惹得毛病。这是一道结构体的题目（不用结构体也能做）。下面上题目。

Milk

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

22Yili 10 500Mengniu 20 10004Yili 10 500Mengniu 20 1000Guangming 1 199Yanpai 40 10000

Sample Output

MengniuMengniu
Hint
In 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

 

题目大意就是让给你牛奶的品牌，花费的钱，体积。让你判断买哪个最合算。注意几点：体积小于两百的不喝，每天喝的是两百ML，牛奶最多喝五千。下面上代码。附带详细的解析。

#include <stdio.h>#include <string.h>struct node   //结构体里面的信息{char name[105];    // 这是牛奶品牌的名字int v;   //    牛奶的体积int money;//    花费的钱} f[105]; //   用来存放以上三个变量void solve()   {int n,m,i,j,p,max;   //定义变量  p表示喝的天数，n代表有几个案例,m代表m个品牌牛奶的名字、价格、体积while(scanf("%d",&n)!=EOF)  {while(n--)  {float sum;// 注意：这里必须用folat 或者double因为（17/5 19/5的值是一样的，所以必须用浮点数，来保证每天喝的价格是不同）float min=999999999;//  让它等于一个很大的数，以便来找最小数。sum=0;j=0;max=-1;//   让max等于一个很小的数，以便来找最大数。scanf("%d",&m);//   输入for(i=1;i<=m;i++)  scanf("%s%d%d",f[i].name,&f[i].money,&f[i].v);//  输入牛奶品牌，花的钱，体积、for(i=1;i<=m;i++){if(f[i].v<200)//  如果小于两百则继续（小于两百就不喝了）continue;     if(f[i].v>1000)//  如果大于一千 则让P=5，因为最多喝五天。p=5;//elsep=f[i].v/200;//     否则除以两百就是喝的天数sum=float(f[i].money/p);//   注意，我就是因为这里WA了许多次。（不能写成float(f[i].money）/p；这样得到的数是整数！！）if(sum<min)//  找出最小的每天喝的价格{j=i;//   把变量赋值给j，就是最小价格的品牌名。min=sum; }else if(sum==min)//   如果价格相同{if(f[i].v>f[j].v)//  则寻找体积大的奶。{j=i;//把变量赋值给j，就是体积大的品牌名}}}printf("%s\n",f[j].name);//  输出}}}int main(){solve();//  解决问题。return 0;}