http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/B

Description

Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.

Codeforces scores are computed as follows: If the maximum point value of a problem is x, and Kevin submitted correctly at minute mbut made w wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.

All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.

Input

The first line of the input contains five space-separated integers m1, m2, m3, m4, m5, where mi (0 ≤ mi ≤ 119) is the time of Kevin's last submission for problem i. His last submission is always correct and gets accepted.

The second line contains five space-separated integers w1, w2, w3, w4, w5, where wi (0 ≤ wi ≤ 10) is Kevin's number of wrong submissions on problem i.

The last line contains two space-separated integers hs and hu (0 ≤ hs, hu ≤ 20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.

Output

Print a single integer, the value of Kevin's final score.

Sample Input

Input

20 40 60 80 100
0 1 2 3 4
1 0

Output

4900

Input

119 119 119 119 119
0 0 0 0 0
10 0

Output

4930

Hint

In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets  of the points on each problem. So his score from solving problems is. Adding in 10·100 = 1000points from hacks, his total score becomes 3930 + 1000 = 4930.

简单模拟。

#include<stdio.h>
#include<algorithm>
#define maxx 10
using namespace std;
long long hu,hs;
double x[maxx]={, , , , },w[maxx],score,m[maxx];
int main(){
for(int i=;i<;i++)
scanf("%lf",&m[i]);
for(int i=;i<;i++)
scanf("%lf",&w[i]);
scanf("%lld%lld",&hs,&hu);
for(int i=;i<;i++)
score+=max(0.3*x[i],(-m[i]/)*x[i]-*w[i]);
score+=*hs-*hu;
printf("%.0f\n",score);//.0f实际上是四舍五入,但是题目保证了答案一定是整数
return ;
}

或者

#include<stdio.h>
#include<algorithm>
#define maxx 10
using namespace std;
long long x[maxx]={, , , ,},w[maxx],score,m[maxx],hu,hs;
int main(){
for(int i=;i<;i++)
scanf("%lld",&m[i]);
for(int i=;i<;i++)
scanf("%ldd",&w[i]);
scanf("%lld%lld",&hs,&hu);
for(int i=;i<;i++)
score+=max(x[i]*/,(x[i]-m[i]*x[i]/)-*w[i]);
score+=*hs-*hu;
printf("%lld\n",score);
return ;
}

  

【 CodeForces 604A】B - 特别水的题2-Uncowed Forces的更多相关文章

  1. 【CodeForces 596A】E - 特别水的题5-Wilbur and Swimming Pool

    Description After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the ...

  2. 【CodeForces 599A】D - 特别水的题4- Patrick and Shopping

    Description  meter long road between his house and the first shop and a d2 meter long road between h ...

  3. 【CodeForces 606A】A -特别水的题1-Magic Spheres

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/A Description Carl is a beginne ...

  4. 【CodeForces 602A】C - 特别水的题3-Two Bases

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/C Description After seeing the ...

  5. Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题

    A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...

  6. Codeforces Round #378 (Div. 2) D题(data structure)解题报告

    题目地址 先简单的总结一下这次CF,前两道题非常的水,可是第一题又是因为自己想的不够周到而被Hack了一次(或许也应该感谢这个hack我的人,使我没有最后在赛后测试中WA).做到C题时看到题目情况非常 ...

  7. CodeForces.158A Next Round (水模拟)

    CodeForces.158A Next Round (水模拟) 题意分析 校赛水题的英文版,坑点就是要求为正数. 代码总览 #include <iostream> #include &l ...

  8. Codeforces Round #612 (Div. 2) 前四题题解

    这场比赛的出题人挺有意思,全部magic成了青色. 还有题目中的图片特别有趣. 晚上没打,开virtual contest打的,就会前三道,我太菜了. 最后看着题解补了第四道. 比赛传送门 A. An ...

  9. Codeforces 828B Black Square(简单题)

    Codeforces 828B Black Square(简单题) Description Polycarp has a checkered sheet of paper of size n × m. ...

随机推荐

  1. UESTC 916 方老师的分身III --拓扑排序

    做法: 如果有a<b的关系,则连一条a->b的有向边,连好所有边后,找入度为0的点作为起点,将其赋为最小的价值888,然后其所有能到的端点,价值加1,加入队列,删去上一个点,然后循环往复, ...

  2. JMeter学习(二)录制脚本

    ---------------------------------------------------------------------------------------------------- ...

  3. 护眼色的RGB值

    网上流行护眼色的RGB值和颜色代码 在搜索引擎搜“护眼色”,就会搜出一堆关于保护眼睛的屏幕颜色文章,说的统统是一种颜色,有点像绿豆沙的颜色.方法就是在屏幕设置里, 色调:85:饱和度:123:亮度:2 ...

  4. mysql高可用方案总结性说明

    MySQL的各种高可用方案,大多是基于以下几种基础来部署的(也可参考:Mysql优化系列(0)--总结性梳理   该文后面有提到)1)基于主从复制:2)基于Galera协议(PXC):3)基于NDB引 ...

  5. setAttribute改变属性,动态改变类

    <style type="text/css"> .box{color:red;} </style> <div>通过setAttribute添加d ...

  6. Bitbucket免费的私有仓库

    1.官网 https://bitbucket.org/ 2.介绍 知乎:http://www.zhihu.com/question/20053312 建议同时用Bitbucket和Github,理由如 ...

  7. SQL 时间处理

    1.获取当前时间 GetDate() 2.获取当前年.月.日 DATEPART(yyyy,GetDate()).DATEPART(m,GetDate()).DATEPART(d,GetDate()) ...

  8. C++引用和java引用的区别

    在c++里的引用其实是一个变量的别名,而java则是一个变量存储实际对象的地址和C++指针很相似

  9. 使用lftp传输文件的shell脚本

    学习参考用,需要服务器上安装lftp. #!/bin/bash #date filepath=/usr/hadoop/bigdata/filterurl filtercount=$(ls $filep ...

  10. 转载:有关SQL server connection Keep Alive 的FAQ(3)

    转载:http://blogs.msdn.com/b/apgcdsd/archive/2012/06/07/sql-server-connection-keep-alive-faq-3.aspx 这个 ...