【51.27%】【codeforces 604A】Uncowed Forces
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
- 【 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 ...
- 【 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 ...
- 【27.91%】【codeforces 734E】Anton and Tree
time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【27.85%】【codeforces 743D】Chloe and pleasant prizes
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【27.66%】【codeforces 592D】Super M
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【27.40%】【codeforces 599D】Spongebob and Squares
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【20.51%】【codeforces 610D】Vika and Segments
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【 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 ...
- 【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
- Finance and Good Society
Finance is a technology of great power. It plays an important role in our society which range from t ...
- 【剑指Offer学习】【面试题49:把字符串转换成整数】
题目:实现一个函数stringToInt,实现把字符串转换成整数这个功能.不能使用atoi或者其它相似的库函数. 题目解析 这看起来是非常easy的题目,实现基本功能 ,大部分人都能用10行之内的代码 ...
- Qt程序调试之Q_ASSERT断言(它是一个宏,接受布尔值,当其中的布尔值为真时,便什么也不做)
在使用Qt开发大型软件时,难免要调试程序,以确保程序内的运算结果符合我们的预期.在不符合预期结果时,就直接将程序断下,以便我们修改. 这就用到了Qt中的调试断言 - Q_ASSERT. 用一个小例子来 ...
- 漫话C++之string字符串类的使用(有汇编分析)
C++中并不提倡继续使用C风格的字符串,而是为字符串定义了专门的类,名为string. 使用前的准备工作 在使用string类型时,需要包含string头文件,且string位于std命名空间内: # ...
- 35.Node.js GET/POST请求
转自:http://www.runoob.com/nodejs/nodejs-module-system.html 在很多场景中,我们的服务器都需要跟用户的浏览器打交道,如表单提交. 表单提交到服务器 ...
- 4.使用 WSDL 指定的标准 SOAP 消息格式
转自:https://technet.microsoft.com/zh-cn/sysinternals/x2ccke44(v=vs.94) 为 XML 文档(定义 Web 服务)定义架构的行业标准 W ...
- 1.25 Python知识进阶 - 封装
封装 示例代码: class Role(object): count = 0 def __init__(self,name,role,weapon,life_value=100,money=15000 ...
- Vue框架学习笔记
<div id="app"> </div> var app = new Vue({ el:"#app", // 绑定的元素 data:{ ...
- Git安装及密钥的生成并上传本地文件到GitHub上
之前用的GitHub,不太熟练,一直在上传的过程中遇到了一些问题,看了网上诸多教程,总觉得很乱,特参考一些资料,总结了一篇完整的操作步骤,从下载安装到上传文件,亲测有效 1.下载Git软件:https ...
- Spark源代码分析之中的一个:Job提交执行总流程概述
Spark是一个基于内存的分布式计算框架.执行在其上的应用程序,依照Action被划分为一个个Job.而Job提交执行的总流程.大致分为两个阶段: 1.Stage划分与提交 (1)Job依照RDD之间 ...