/*

 L (>=60), the lower bound of the qualified grades --
that is, only the ones whose grades of talent and virtue are both not below
this line will be ranked; and H (<100), the higher line of qualification ,those with both grades not below this line are considered as the "sages",
and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as
the "noblemen", and are also ranked in non-increasing order according to
their total grades, but they are listed after the "sages". Those with both grades below H,
but with virtue not lower than talent are considered as the "fool men".
They are ranked in the same way but after the "noblemen". The rest of people whose grades both pass the L line are ranked
after the "fool men". */ #include <string.h>
#include <algorithm>
#include <vector>
#include <stdio.h>
using namespace std; struct peo
{
char num[];
int all,t,v;
int id;
}; bool cmp(peo a,peo b)
{
if(a.id==b.id)
{
if(a.all == b.all)
{
if(a.v == b.v )
return (strcmp(a.num,b.num)<);
else return a.v > b.v;
}
else return a.all > b.all;
}
else return a.id<b.id; } int main()
{ int i,n,low,high,v,t;
char num[];
while(scanf("%d%d%d",&n,&low,&high)!=EOF)
{ vector<peo> VP;
for(i=;i<n;i++)
{
getchar();
scanf("%s %d %d",num,&v,&t);
if(t>=low && v>=low)
{
peo pp;
strcpy(pp.num,num);
pp.v=v;
pp.t=t;
pp.all=v+t;
if(v>=high && t>=high)
pp.id=;
else if(v>= high && t<high)
pp.id=;
else if(v< high && t< high && v>=t)
pp.id=;
else
pp.id=; VP.push_back(pp);
}
} sort(VP.begin(),VP.end(),cmp);
printf("%d\n",VP.size());
for(i=;i<VP.size();i++)
printf("%s %d %d\n",VP[i].num,VP[i].v,VP[i].t); }
return ;
}

1062 Talent and Virtue (25)的更多相关文章

  1. 1062. Talent and Virtue (25)【排序】——PAT (Advanced Level) Practise

    题目信息 1062. Talent and Virtue (25) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B About 900 years ago, a Chine ...

  2. PAT 甲级 1062 Talent and Virtue (25 分)(简单,结构体排序)

    1062 Talent and Virtue (25 分)   About 900 years ago, a Chinese philosopher Sima Guang wrote a histor ...

  3. 1062 Talent and Virtue (25分)(水)

    About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about ...

  4. pat 1062. Talent and Virtue (25)

    难得的一次ac 题目意思直接,方法就是对virtue talent得分进行判断其归属类型,用0 1 2 3 4 表示 不合格 sage noblemen foolmen foolmen 再对序列进行排 ...

  5. PAT (Advanced Level) 1062. Talent and Virtue (25)

    简单排序.题意较长. #include<cstdio> #include<cstring> #include<cmath> #include<queue> ...

  6. PAT甲题题解-1062. Talent and Virtue (25)-排序水题

    水题,分组排序即可. #include <iostream> #include <cstdio> #include <algorithm> #include < ...

  7. 【PAT甲级】1062 Talent and Virtue (25 分)

    题意: 输入三个正整数N,L,H(N<=1E5,L>=60,H<100,H>L),分别代表人数,及格线和高水平线.接着输入N行数据,每行包括一个人的ID,道德数值和才能数值.一 ...

  8. pat1062. Talent and Virtue (25)

    1062. Talent and Virtue (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Li Abou ...

  9. 1062 Talent and Virtue (25 分)

    1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a history ...

随机推荐

  1. 使用apache和IIS,共用80端口的一个解决方案【转】

    将apache设为使用80端口,IIS使用其它端口,比如81,然后将apache作为IIS的代理. 在httpd.conf里面,取消下面四行的注释: LoadModule proxy_module m ...

  2. CentOS(九)--与Linux文件和目录管理相关的一些重要命令①

       接上一篇文章,实际生产过程中的目录管理一定要注意用户是root 还是其他用户. 一.目录与路径 1.相对路径与绝对路径 因为我们在Linux系统中,常常要涉及到目录的切换,所以我们必须要了解 & ...

  3. wap测试学习

    注意要点 UI元素 修改源码 物理键操作(回车.确认) 焦点 习惯性操作(前进.后退.屏幕翻转和停止) 刷新 重启服务器 重启浏览器 异常关闭 书签 cookies/session 缓存 接口 URL ...

  4. 【转载】Android推送方案分析(MQTT/XMPP/GCM)

    http://m.oschina.net/blog/82059 本文主旨在于,对目前Android平台上最主流的几种消息推送方案进行分析和对比,比较客观地反映出这些推送方案的优缺点,帮助大家选择最合适 ...

  5. Java Servlet-http协议

    ---恢复内容开始--- 互联网三大基石: url:定位数据 html:显示数据 http:传输数据

  6. Rebind and Rewind in Execution Plans

    http://www.scarydba.com/2011/06/15/rebind-and-rewind-in-execution-plans/ Ever looked at an execution ...

  7. CentOS安装Node.js简单教程

    记录一下自己安装过程 先安装gcc-c++编译环境和openssl  代码如下 复制代码 yum install gcc-c++ openssl-devel wget http://nodejs.or ...

  8. Swift字典类

    在Foundation框架中提供一种字典集合,它是由“键-值”对构成的集合.键集合不能重复,值集合没有特殊要求.键和值集合中的元素可以是任何对象,但是不能是nil.Foundation框架字典类也分为 ...

  9. The influence of informal governance mechanisms on knowledge integration

    Title:The influence of informal governance mechanisms on knowledge integration within cross-function ...

  10. C#消息模拟

    C#中消息的工作流程: C#中的消息被Application类从应用程序消息队列中取出,然后分发到消息对应的窗体,窗体对象的第一个响应函数是对象中的protected override void Wn ...