time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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 m but 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.

Examples

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

Note

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 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930.

【题目链接】:http://codeforces.com/contest/604/problem/A

【题解】



不要按照样例解释的方法算。。

总感觉那个方法是误导的。。

直接按照所给的方法算就好了。

有除法、还是用double的吧.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x) typedef pair<int,int> pii;
typedef pair<LL,LL> pll; //const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0); double m[6],w[6],hs,hu;
double poi[6]; int main()
{
//freopen("F:\\rush.txt","r",stdin);
poi[1] = 500,poi[2] = 1000,poi[3] = 1500,poi[4] = 2000,poi[5] = 2500;
rep1(i,1,5)
cin >> m[i];
rep1(i,1,5)
cin >> w[i];
cin >> hs >> hu;
rep1(i,1,5)
{
double temp1 = 0.3*poi[i];
double temp2 = (1-m[i]/250)*poi[i]-50*w[i];
poi[i] = max(temp1,temp2);
}
double ans = 0;
rep1(i,1,5)
ans+=poi[i];
ans += (hs*100-50*hu);
printf("%.0lf\n",ans);
return 0;
}

【51.27%】【codeforces 604A】Uncowed Forces的更多相关文章

  1. 【 CodeForces 604A】B - 特别水的题2-Uncowed Forces

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/B Description Kevin Sun has jus ...

  2. 【 BowWow and the Timetable CodeForces - 1204A 】【思维】

    题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...

  3. 【27.91%】【codeforces 734E】Anton and Tree

    time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  4. 【27.85%】【codeforces 743D】Chloe and pleasant prizes

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  5. 【27.66%】【codeforces 592D】Super M

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  6. 【27.40%】【codeforces 599D】Spongebob and Squares

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  7. 【20.51%】【codeforces 610D】Vika and Segments

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  8. 【 Educational Codeforces Round 51 (Rated for Div. 2) F】The Shortest Statement

    [链接] 我是链接,点我呀:) [题意] [题解] 先处理出来任意一棵树. 然后把不是树上的边处理出来 对于每一条非树边的点(最多21*2个点) 在原图上,做dijkstra 这样就能处理出来这些非树 ...

  9. 【47.40%】【codeforces 743B】Chloe and the sequence

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

随机推荐

  1. 基于 Cookie 的 SSO 中间件 kisso

    kisso  =  cookie sso 基于 Cookie 的 SSO 中间件,它是一把快速开发 java Web 登录系统(SSO)的瑞士军刀.欢迎大家使用 kisso !! kisso 帮助文档 ...

  2. cocos2dx——lua自己主动和手动绑定

    [自己主动绑定] 參考:http://my.oschina.net/skyhacker2/blog/298397 主要是通过引擎自带的tools/tolua,主要过程例如以下: 1.编写好要导出的c+ ...

  3. windows 2016 配置 VNC 服务

    windows 2016 配置 VNC 服务 下载windows版 https://www.realvnc.com/download/vnc/ 安装时勾选 vncserver 进入 "C:\ ...

  4. ip地址个数的计算

    一个IP地址,却关联太多的知识 二进制与 8 比特 电脑中显示出来的数字是 10 进制的,键盘的每一个键都由一个 8 位的二进制编码,所以 1 字节等于 8 比特.对数字而言,1 的二进制是 0000 ...

  5. 类名引用static变量好处

    不仅强调了变量static的结构,而且在有些情况下他还为编译器进行优化提供了更好的机会.

  6. carousel轮播器

    <div id="myCarousel" class="carousel slide" data-ride="carousel" st ...

  7. 开发板Ping不通虚拟机和主机

    Ubuntu 16.04      win7 笔记本连接学校的无线网 开发板S3c2440与笔记本仅通过COM连接 问题描述: 设置了桥接,主机与虚拟机IP在同一网段后,主机与虚拟机可以Ping,但是 ...

  8. SQLITE数据表主键设置Id自增方法

    SQLITE数据表主键设置Id自增方法 标签: sqliteintegerinsertnulltableapi 2010-01-12 08:39 35135人阅读 评论(8) 收藏 举报  分类: S ...

  9. 导出查询结果到csv文件

    set colsep ,   set feedback off   set heading off   set trimout on   spool my.csv  select * from emp ...

  10. 自己写的Android图表库XCL-Charts一些旧的样例

    话说有了灵感就要抓住,来了兴趣就要去研究它. 所以尽管近期非常忙.但我还是没有丢下Android图表实现的研究.最终如今我的图表库基类 基本上已经有点模样了.不在是小打小闹,而是能依传入參数非常灵活的 ...