hdu 1070(结构体排序)
来源:互联网 发布:js清空input file内容 编辑:程序博客网 时间:2024/05/17 04:33
Milk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13639 Accepted Submission(s): 3328
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.
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.
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
MengniuMengniuHintIn 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
难点:关于sort排序的不能得心应手
时间:8.8
做题人:洛可可
主要考查的是结构体的二级排序,并且是double 型的数据不能直接 等号表示,两者相等,因为牵扯到精度的问题。
二级排序的sort代码如下:
<span style="font-size:14px;">#include<stdio.h>#include<algorithm>using namespace std;struct milk{char log[110];int p;int v;double xjb;}a[110];int cmp(milk x,milk y){if(x.xjb!=y.xjb) return x.xjb<y.xjb;return x.v>y.v;}int main(){int n,i,m;scanf("%d",&n);while(n--){scanf("%d",&m);for(i=0;i<m;i++){scanf("%s%d%d",a[i].log,&a[i].p,&a[i].v);//getchar();if(a[i].v>1200)a[i].xjb=a[i].p*1.0/1200;elsea[i].xjb=a[i].p*1.0/a[i].v;}sort(a,a+m,cmp);for(i=0;i<m;i++){if(a[i].v<200)continue;else{printf("%s\n",a[i].log);break;}}}return 0;}</span>
<span style="font-size:14px;">#include<stdio.h>#include<stdlib.h>#include<string.h>struct milk{char log[110];int p;int v;double xjb;}a[110];int cmp(const void *a,const void *b){if((*(milk *)a).xjb>(*(milk *)b).xjb)return 1;else if(((*(milk *)a).xjb<(*(milk *)b).xjb))return -1;elsereturn (((milk *)a)->v)<(((milk *)b)->v);}int main(){int t,i,n;scanf("%d",&t);while(t--){scanf("%d",&n);getchar();for(i=0;i<n;i++){scanf("%s %d %d",a[i].log,&a[i].p,&a[i].v);getchar();if(a[i].v>1200)a[i].xjb=a[i].p*1.0/1200;</span>
<span style="font-size:14px;"> elsea[i].xjb=a[i].p*1.0/a[i].v;}qsort(a,n,sizeof(a[0]),cmp);for(i=0;i<n;i++){if(a[i].v<200)continue;else{printf("%s\n",a[i].log);break;}}}return 0;}</span>
0 0
- hdu 1070(结构体排序)
- HDU-1070 Milk 结构体排序
- hdu 1785 结构体排序
- HDU 1236 结构体排序
- hdu 1263 结构体排序
- hdu _1070_结构体排序
- HDU 2093 考试排名---结构体排序
- hdu 4310 Hero(贪心 结构体排序)
- HDU 1263 水果(结构体排序)
- HDU 1236 排名 (排序+结构体)
- hdu 2093 考试排名(结构体排序)
- 【结构体排序】hdu 2409 Team Arrangement
- Hdu 2093(三级结构体排序)
- 【HDU 5499】+ sort 结构体排序
- hdu Problem-5702(结构体排序)
- HDU 2093 考试排名【结构体排序】
- HDU 1862 EXCEL排序(结构体排序)
- HDU--杭电--1177--"Accepted today?"--结构体排序
- UML符号---设计模式之基础
- uploadify 导致Chrome频繁崩溃Crash
- 本地通知——UILocalNotification
- hdoj 2030 汉字统计 氺过~
- sql触发器,函数
- hdu 1070(结构体排序)
- error C2041: illegal digit '#' for base '8'
- Linux多线程同步机制
- 在TOMCAT下配置工程的默认访问
- 项目管理------项目度的控制
- Echart饼图、柱状图、折线图(pie、bar、line)添加点击事件
- Unity3D-RPG项目实战(1):引擎特殊目录
- 转载:Android HttpClient & HttpUrlconnection
- HDU 1251 统计难题(字典树)