ZCC Loves Codefires

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 790    Accepted Submission(s): 420

Problem Description
Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137".
It was on Codefires(CF), an online competitive programming site, that ZCC knew Memset137, and immediately became his fan.
But why?
Because Memset137 can solve all problem in rounds, without
unsuccessful submissions; his estimation of time to solve certain
problem is so accurate, that he can surely get an Accepted the second he
has predicted. He soon became IGM, the best title of Codefires.
Besides, he is famous for his coding speed and the achievement in the
field of Data Structures.
After become IGM, Memset137 has a new goal: He wants his score in CF rounds to be as large as possible.
What is score? In Codefires, every problem has 2 attributes, let's
call them Ki and Bi(Ki, Bi>0). if Memset137 solves the problem at Ti-th
second, he gained Bi-Ki*Ti score. It's guaranteed Bi-Ki*Ti is always
positive during the round time.
Now that Memset137 can solve every
problem, in this problem, Bi is of no concern. Please write a program
to calculate the minimal score he will lose.(that is, the sum of
Ki*Ti).
 
Input
The first line contains an integer N(1≤N≤10^5), the number of problem in the round.
The second line contains N integers Ei(1≤Ei≤10^4), the time(second) to solve the i-th problem.
The last line contains N integers Ki(1≤Ki≤10^4), as was described.
 
Output
One integer L, the minimal score he will lose.
 
Sample Input
3
10 10 20
1 2 3
 
Sample Output
150

Hint

Memset137 takes the first 10 seconds to solve problem B, then solves problem C at the end of the 30th second. Memset137 gets AK at the end of the 40th second.
L = 10 * 2 + (10+20) * 3 + (10+20+10) * 1 = 150.

 
Author
镇海中学
 
Source
代码:
 #include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
int e,t;
bool operator < (const node a) const{
return e*a.t>a.e*t;
}
}map[];
int main(){
int n,i;
__int64 tat,ans;
while(scanf("%d",&n)!=EOF){
for(i=;i<n;i++)
scanf("%d",&map[i].t);
for(i=;i<n;i++)
scanf("%d",&map[i].e);
sort(map,map+n);
for(ans=tat=i=;i<n;i++){
tat+=map[i].t;
ans+=tat*map[i].e;
}
printf("%I64d\n",ans);
}
return ;
}

2014---多校训练2(ZCC Loves Codefires)的更多相关文章

  1. 2014 (多校)1011 ZCC Loves Codefires

    自从做了多校,整个人都不好了,老是被高中生就算了,题老是都不懂=-=原谅我是个菜鸟,原谅我智力不行.唯一的水题. Problem Description Though ZCC has many Fan ...

  2. HDU 4882 ZCC Loves Codefires(贪心)

     ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/O ...

  3. hdu 4882 ZCC Loves Codefires(数学题+贪心)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4882 ------------------------------------------------ ...

  4. HDU-4882 ZCC Loves Codefires

    http://acm.hdu.edu.cn/showproblem.php?pid=4882 ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Ot ...

  5. HDU 4882 ZCC Loves Codefires (贪心)

    ZCC Loves Codefires 题目链接: http://acm.hust.edu.cn/vjudge/contest/121349#problem/B Description Though ...

  6. 2014多校第二场1011 || HDU 4882 ZCC Loves Codefires (贪心)

    题目链接 题意 : 给出n个问题,每个问题有两个参数,一个ei(所要耗费的时间),一个ki(能得到的score).每道问题需要耗费:(当前耗费的时间)*ki,问怎样组合问题的处理顺序可以使得耗费达到最 ...

  7. BZOJ 3850: ZCC Loves Codefires【贪心】

    Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137". It wa ...

  8. 【BZOJ】3850: ZCC Loves Codefires(300T就这样献给了水题TAT)

    http://www.lydsy.com/JudgeOnline/problem.php?id=3850 题意:类似国王游戏....无意义.. #include <cstdio> #inc ...

  9. 【BZOJ】【3850】ZCC Loves Codefires

    贪心 就跟NOIP2012国王游戏差不多,考虑交换相邻两题的位置,对其他题是毫无影响的,然后看两题顺序先后哪个更优.sort即可. WA了一次的原因:虽然ans开的是long long,但是在这一句: ...

随机推荐

  1. jquery相对选择器,又叫context选择器,上下文选择器;find()与children()区别

    jquery相对选择器有两个参数,jQuery函数的第二个参数可以指定DOM元素的搜索范围(即以第二个参数指定的内容为容器查找指定元素). 第二个参数的不同的类型,对应的用法如下表所示. 类型 用法 ...

  2. ps aux 查看进程信息

    [root@localhost Desktop]# ps auxUSER PID %CPU %MEM VSZ RSS TTY STAT START TIME COMMANDroot 1 0.0 0.3 ...

  3. Trigger Execution Sequence Of Oracle Forms

    Sequence of triggers fires on Commit.1.  KEY Commit2.  Pre Commit3.  Pre/On/Post Delete4.  Pre/On/Po ...

  4. The CLR's Execution Model

    the native code generator tool:NGen.exe optimization tool:MPGO.exe 所有类型最终都继承自System.Object.则所有类型都有如下 ...

  5. CUBRID学习笔记 7 ms常见错误

    基本不是权限问题,就是dll问题.  重新下载或应用dll注意版本. 权限的问题,先本机测试. 看看在web管理有无问题.  剩下的基本就简单了 欢迎转载 ,转载时请保留作者信息.本文版权归本人所有, ...

  6. FZU 2215 Simple Polynomial Problem(简单多项式问题)

    Description 题目描述 You are given an polynomial of x consisting of only addition marks, multiplication ...

  7. android tablelayout 显示图片

    当在tablelayout中显示图片时,设置imageView为固定大小时,会出现divide by zero 错误 将LayoutParams 改为 TableRow.LayoutParams即可 ...

  8. CISCO VPN出现网关报错

    今天尝试使用发现报错: 重启VPN服务即可

  9. HDU 1754

    成段更新 easy #include <stdio.h> #include <string.h> #include <math.h> #include <io ...

  10. sqlserver 批量删除存储过程(转)

    sqlserver一次只能删除一个存储过程,如果多了,需要很长时间才能删完,所以写了一段语句,直接就把当然数据库下所有用户自定义的存储过程给drop了.不过使用都请留心,当前打开的数据库哦.下面贴代码 ...