简单模拟题。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<map>
#include<queue>
using namespace std; int cost[];
int n;
struct X
{
string name;
int dd,hh,mm,len,f;
}s[],ans[];
int month;
bool flag[]; bool cmp(const X&a,const X&b)
{
if(a.len==b.len) return a.f>b.f;
return a.len<b.len;
} bool cmp2(const X&a,const X&b)
{
if(a.name==b.name) return a.len<b.len;
return a.name<b.name;
} int get(int a,int b)
{
int sum=; for(int i=ans[a].len;i<ans[b].len;i++)
{
int DD=i/(*);
int HH=(i-DD**)/;
sum=sum+cost[HH];
}
return sum;
} int main()
{
for(int i=;i<;i++) scanf("%d",&cost[i]);
scanf("%d",&n);
for(int i=;i<=n;i++)
{
cin>>s[i].name;
scanf("%d:%d:%d:%d",&month,&s[i].dd,&s[i].hh,&s[i].mm);
s[i].len=s[i].dd**+s[i].hh*+s[i].mm;
char op[]; scanf("%s",op);
if(op[]=='n') s[i].f=;
else s[i].f=;
}
memset(flag,,sizeof flag); sort(s+,s++n,cmp);
int tot=;
for(int i=;i<=n;i++)
{
if(flag[i]==) continue;
if(s[i].f==) continue;
for(int j=i+;j<=n;j++)
{ if(s[i].name==s[j].name)
{
if(s[j].f==&&flag[j]==){
ans[++tot]=s[i];
ans[++tot]=s[j];
flag[i]=; flag[j]=;
}
break;
}
}
} sort(ans+,ans++tot,cmp2); int pos=;
cout<<ans[].name<<" "; printf("%02d\n",month);
printf("%02d:%02d:%02d ",ans[pos].dd,ans[pos].hh,ans[pos].mm);
printf("%02d:%02d:%02d ",ans[pos+].dd,ans[pos+].hh,ans[pos+].mm);
printf("%d ",ans[pos+].len-ans[pos].len);
printf("$%.2lf\n",1.0*get(,)/); double sum=1.0*get(,)/; pos=pos+; for(int i=pos;i<=tot;i=i+)
{
if(ans[i].name==ans[i-].name)
{
printf("%02d:%02d:%02d ",ans[i].dd,ans[i].hh,ans[i].mm);
printf("%02d:%02d:%02d ",ans[i+].dd,ans[i+].hh,ans[i+].mm);
printf("%d ",ans[i+].len-ans[i].len);
printf("$%.2lf\n",1.0*get(i,i+)/);
sum=sum+1.0*get(i,i+)/;
}
else
{
printf("Total amount: $%.2lf\n",sum);
cout<<ans[i].name<<" "; printf("%02d\n",month);
printf("%02d:%02d:%02d ",ans[i].dd,ans[i].hh,ans[i].mm);
printf("%02d:%02d:%02d ",ans[i+].dd,ans[i+].hh,ans[i+].mm);
printf("%d ",ans[i+].len-ans[i].len);
printf("$%.2lf\n",1.0*get(i,i+)/);
sum=1.0*get(i,i+)/;
}
}
printf("Total amount: $%.2lf\n",sum);
return ;
}

PAT (Advanced Level) 1016. Phone Bills (25)的更多相关文章

  1. PTA (Advanced Level) 1016 Phone Bills

    Phone Bills A long-distance telephone company charges its customers by the following rules: Making a ...

  2. PAT (Advanced Level) 1114. Family Property (25)

    简单DFS. #include<cstdio> #include<cstring> #include<cmath> #include<vector> # ...

  3. PAT (Advanced Level) 1109. Group Photo (25)

    简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #i ...

  4. PAT (Advanced Level) 1105. Spiral Matrix (25)

    简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<map> #incl ...

  5. PAT (Advanced Level) 1101. Quick Sort (25)

    树状数组+离散化 #include<cstdio> #include<cstring> #include<cmath> #include<map> #i ...

  6. PAT (Advanced Level) 1071. Speech Patterns (25)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  7. PAT (Advanced Level) 1063. Set Similarity (25)

    读入之后先排序. 询问的时候可以o(m)效率得到答案. #include<cstdio> #include<cstring> #include<cmath> #in ...

  8. PAT (Advanced Level) 1059. Prime Factors (25)

    素因子分解. #include<iostream> #include<cstring> #include<cmath> #include<algorithm& ...

  9. PAT (Advanced Level) 1051. Pop Sequence (25)

    简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...

随机推荐

  1. JavaScript 伪造 Referer 来路方法

    Javascript 是一种由Netscape的LiveScript发展而来的原型化继承的基于对象的动态类型的区分大小写的客户端脚本语言,主要目的是为了解决服务器端语言,比如Perl,遗留的速度问题, ...

  2. 剑指offer 旋转数组

    class Solution { public: int minNumberInRotateArray(vector<int> rotateArray) { //常规的遍历方法时间是O(N ...

  3. 在调试安卓系统的时候需要这个 ”adb disable-verity“

    在调试设备的时候.想要对文件进行读写 于是使用adb remount 出现提示. 请使用 ”adb  disable-verity“ 于是使用adb  disable-verity 的命令. 得到如下 ...

  4. 【转】linux ls -l的详解

    原文:http://blog.csdn.net/sjzs5590/article/details/8254527 以root的家目录为例: 可以看到,用ls -l命令查看某一个目录会得到一个7个字段的 ...

  5. Food on the Plane

    Food on the Plane time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  6. WPF Template模版之DataTemplate与ControlTemplate的关系和应用【二】

    1. DataTemplate和ControlTemplate的关系 学习过DataTemplate和ControlTemplate,你应该已经体会到,控件只是数据的行为和载体,是个抽象的概念,至于它 ...

  7. OpenGL---------BMP文件格式

    计算机保存图象的方法通常有两种:一是“矢量图”,一是“像素图”.矢量图保存了图象中每一几何物体的位置.形状.大小等信息,在显示图象时,根据这些信息计算得到完整的图象.“像素图”是将完整的图象纵横分为若 ...

  8. /var/lib/mysql/mysql.sock错误的解决办法

    问题描述: 使用mysql -uroot -p登录出现找不到 /var/lib/mysql/mysql.sock或者被使用的问题. 可以用如下命令登录:mysql -p --socket=/tmp/m ...

  9. jquery 左边分类+插件

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  10. Android与路由器连接服务

    界面UI: package my.work.Library; import java.util.Timer; import java.util.TimerTask; import java.util. ...