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. GZFramwork数据库层《一》普通表增删改查

    运行结果:     使用代码生成器(GZCodeGenerate)生成tb_MyUser的Model 生成器源代码下载地址: https://github.com/GarsonZhang/GZCode ...

  2. [SAP ABAP开发技术总结]以二进制、字符模式下载文件

    声明:原创作品,转载时请注明文章来自SAP师太技术博客( 博/客/园www.cnblogs.com):www.cnblogs.com/jiangzhengjun,并以超链接形式标明文章原始出处,否则将 ...

  3. [SAP ABAP开发技术总结]数据输入输出转换、小数位/单位/货币格式化

    声明:原创作品,转载时请注明文章来自SAP师太技术博客( 博/客/园www.cnblogs.com):www.cnblogs.com/jiangzhengjun,并以超链接形式标明文章原始出处,否则将 ...

  4. [SAP ABAP开发技术总结]CLEAR、REFRESH、FREE内表清理区别

    声明:原创作品,转载时请注明文章来自SAP师太技术博客( 博/客/园www.cnblogs.com):www.cnblogs.com/jiangzhengjun,并以超链接形式标明文章原始出处,否则将 ...

  5. HDU 5818 Joint Stacks(联合栈)

    HDU 5818 Joint Stacks(联合栈) Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Ja ...

  6. C#正则表达式编程(三):Match类和Group类用法

    前面两篇讲述了正则表达式的基础和一些简单的例子,这篇将稍微深入一点探讨一下正则表达式分组,在.NET中正则表达式分组是用Match类来代表的.首先先看一段代码: /// <summary> ...

  7. [转发] 理解 oauth 2.0

    原文: http://www.ruanyifeng.com/blog/2014/05/oauth_2_0.html oauth 的各种编程语言实现: http://oauth.net/2/ 理解OAu ...

  8. O(n)获得中位数及获得第K小(大)的数

    首先,中位数问题可以归结为求 K=n/2的 第K小元素,并无明显区别. 第一种方法,用MaxHeap,大小为K的大顶堆,能够求出最小的K的元素,复杂度为O(n*logK). 当K较大时,复杂度会较高. ...

  9. TCP/IP协议学习(一) LWIP实现网络远程IAP下载更新

    最近需要实现通过TCP/IP远程IAP在线更新功能,忙了2周终于在原有嵌入式服务器的基础上实现了该功能,这里就记录下实现的过程. IAP又称在应用编程,其实说简单点就是实现不需要jlink,仅通过芯片 ...

  10. [CSS] vertical-align

    原文地址: http://www.zhangxinxu.com/wordpress/2010/05/%E6%88%91%E5%AF%B9css-vertical-align%E7%9A%84%E4%B ...