【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
- IOS经常使用的性能优化策略
1.用ARC管理内存 2.对于UITableView使用重用机制 3.UIView及其子类设置opaque=true 4.主进程是用来绘制UI的,所以不要堵塞 5.慎用XIB,由于XIB创建UIVie ...
- Android自定义视图
Android框架为我们提供了大量的视图类来帮助我们做好展示信息以及同用户进行交互的工作.然后有时候,我们的app或许需要一些在Android内建视图之外特殊的视图,那么此时我们就需要自定义视图.下面 ...
- 3、Task.Factory属性
3.Task.Factory属性 Task类提供了一个Factory静态属性,这个属性返回一个TaskFactory对象. Task task = Task.Factory.StartNew(Task ...
- InstallShield详细制作说明(四)
十.编译打包
- sql server还原数据库代码
RESTORE DATABASE ExaminationsystemFROM DISK = 'C:\Users\admin\Desktop\20140324.bak'with replace,MOVE ...
- 【2017中国大学生程序设计竞赛 - 网络选拔赛】A Secret
[链接]http://acm.hdu.edu.cn/showproblem.php?pid=6153 [题意] ,S2中出现的次数与其长度的乘积之和. [题解] 扩展KMP的模板题. 首先,把S2和 ...
- 计科1111-1114班第一次实验作业(NPC问题——回溯算法、聚类分析)
实验课安排 地点: 科技楼423 时间: 计科3-4班---15周周一上午.周二下午 计科1-2班---15周周一下午.周二晚上(晚上时间从18:30-21:10) 请各班学委在实验课前飞信通知大家 ...
- hdu 2795 Billboard(线段树单点更新)
Billboard Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total ...
- menu-普通menu弹出框样式
今天接触到了menu弹出框样式.主要就是在theme下进行调整.现在把接触到的知识点总结一下. 在theme中,跟menu有关的几个属性如下 <item name="panelBack ...
- ORACLE11g R2【RAC+ASM→单实例FS】
ORACLE11g R2[RAC+ASM→单实例FS] 11g R2 RAC+ASMà单实例FS的DG,建议禁用OMF. 本演示案例所用环境: primary standby OS Hostnam ...