Treasure the new start, freshmen!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23047    Accepted Submission(s): 6814

Problem Description
background:
A new semester comes , and the HDU also meets its 50th birthday. No matter what's your major, the only thing I want to tell you is:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school.
"No pain, No Gain", HDU also has scholarship, who can win it? That's mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.
If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA is
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
If there is a 0 <= Si < 60, The GPA is always not existed.
 
Input
The first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).
Notice: There is no blank in the Course Name. All the Inputs are legal
 
Output
Output the GPA of each case as discribed above, if the GPA is not existed, ouput:"Sorry!", else just output the GPA value which is rounded to the 2 digits after the decimal point. There is a blank line between two test cases. 
 
Sample Input
2
3
Algorithm 3 97
DataStruct 3 90
softwareProject 4 85
2
Database 4 59
English 4 81
 
Sample Output
90.10
Sorry!
 #include<bits/stdc++.h>
using namespace std;
char s[];
int main()
{
int t;
while(~scanf("%d",&t))
{
while(t--)
{
int n;
scanf("%d",&n);int r=;
double sum1=,sum2=;
double cre,score;
for(int i=;i<n;i++)
{
memset(s,'\0',sizeof(s));
scanf("%s %lf %lf",s,&cre,&score);
if(score<)r=;
sum1=cre*score+sum1;
sum2=cre+sum2;
}
if(r)
{
printf("Sorry!\n");
}
else
{
printf("%.2lf\n",sum1/sum2);
}
if(t!=)printf("\n");
}
}
return ;
}

hdu2061 Treasure the new start, freshmen!(暴力简单题)的更多相关文章

  1. HDU 6463.超级无敌简单题-卡边界的暴力 (“字节跳动-文远知行杯”广东工业大学第十四届程序设计竞赛)

    超级无敌简单题 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Sub ...

  2. [BZOJ2683][BZOJ4066]简单题

    [BZOJ2683][BZOJ4066]简单题 试题描述 你有一个N*N的棋盘,每个格子内有一个整数,初始时的时候全部为0,现在需要维护两种操作: 命令 参数限制 内容 1 x y A 1<=x ...

  3. 又一道简单题&&Ladygod(两道思维水题)

    Ladygod Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others) Submit S ...

  4. 【BZOJ4066】简单题 KDtree

    [BZOJ4066]简单题 Description 你有一个N*N的棋盘,每个格子内有一个整数,初始时的时候全部为0,现在需要维护两种操作: 命令 参数限制 内容 1 x y A 1<=x,y& ...

  5. [bzoj4066/2683]简单题_KD-Tree

    简单题 bzoj-4066 题目大意:n*n的棋盘,开始为均为0,支持:单点加权值,查询矩阵权值和,强制在线. 注释:$1\le n\le 5\cdot 10^5$,$1\le m \le 2\cdo ...

  6. bzoj3687简单题(dp+bitset优化)

    3687: 简单题 Time Limit: 10 Sec  Memory Limit: 512 MBSubmit: 861  Solved: 399[Submit][Status][Discuss] ...

  7. 树状数组例题-数星星,简单题easy,校门外的树2,清点人数

    [例1]数星星 天空中有一些星星,这些星星都在不同的位置,每个星星都有个坐标,如果一个星星的左下方(包括正左和正下)有k颗星星,就说这颗星星是k级的. 比如,上图中,星星5是3级的(1,2,4在其左下 ...

  8. BZOJ 2683: 简单题

    2683: 简单题 Time Limit: 50 Sec  Memory Limit: 128 MBSubmit: 913  Solved: 379[Submit][Status][Discuss] ...

  9. 【BZOJ-1176&2683】Mokia&简单题 CDQ分治

    1176: [Balkan2007]Mokia Time Limit: 30 Sec  Memory Limit: 162 MBSubmit: 1854  Solved: 821[Submit][St ...

随机推荐

  1. concatenate

    concatenate函数的作用是对numpy数组或tensor进行拼接 >>> a = np.array([[1, 2], [3, 4]]) >>> b = np ...

  2. UESTC - 1987 童心未泯的帆宝和乐爷 (第k短路 A*算法+SPFA算法 模板)

    传送门: http://www.qscoj.cn/#/problem/show/1987 童心未泯的帆宝和乐爷 Edit Time Limit: 10000 MS     Memory Limit: ...

  3. 404 Note Found 队-Alpha5

    目录 组员情况 组员1(组长):胡绪佩 组员2:胡青元 组员3:庄卉 组员4:家灿 组员5:凯琳 组员6:翟丹丹 组员7:何家伟 组员8:政演 组员9:黄鸿杰 组员10:刘一好 组员11:何宇恒 展示 ...

  4. .NET Core中基类可以反射子类的成员

    我们定义一个类DemoA,再定义一个类DemoB继承DemoA.当构造一个DemoB类对象后,我们可以通过其调用基类DemoA中的方法来反射子类DemoB的成员. 新建一个.NET Core控制台项目 ...

  5. POJ 3762 The Bonus Salary!(最小K覆盖)

    POJ 3762 The Bonus Salary! 题目链接 题意:给定一些任务.每一个任务有一个时间,有k天.一个时间仅仅能运行一个任务,每一个任务有一个价值.问怎么安排能得到最多价值 思路:典型 ...

  6. No active profile set, falling back to default profiles: default

    No active profile set, falling back to default profiles: default 这个错误是由于idea没有设置默认启动环境,设置即可

  7. Servlet基础知识总结

    Servlet是JavaWeb应用开发的核心组件.Servlet运行在Servlet容器中(例如最常用的Tomcat),它可以为各种客户请求提供相应服务.Servlet可以轻松完成以下任务: 动态生成 ...

  8. jQuery实现全选、不选和反选功能

    jQuery结合Font Awesome字体图标实现全选.不选和反选功能 Font Awesome字体图标链接地址:http://www.fontawesome.com.cn/faicons/ 效果: ...

  9. 基于 HTML5 WebGL 的 3D 风机 Web 组态工业互联网应用

    基于 HTML5 WebGL 的 3D 风机 Web 组态工业互联网应用 前言 在目前大数据时代背景之下,数据可视化的需求也变得越来越庞大,在数据可视化的背景之下,通过智能机器间的链接并最终将人机链接 ...

  10. js 刷新父页面

    //刷新父页面 window.opener.location.reload();