http://acm.hdu.edu.cn/showproblem.php?pid=2061

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!
 
代码:

#include <bits/stdc++.h>
using namespace std; int main()
{
int T;
char name[50];
double num[1111],score[1111];
scanf("%d",&T);
for(int i=1; i<=T; i++)
{
if(i!=1)
printf("\n");
int x,flag=0;
scanf("%d",&x);
double sum=0,ans=0;
for(int j=1; j<=x; j++)
{
scanf("%s%lf%lf",name,&num[j],&score[j]);
if(score[j]<60)
flag=1;
sum+=num[j]*score[j];
ans+=num[j];
}
//cout<<sum<<" "<<ans<<" "<<flag<<endl;
if(flag==0)
printf("%.2lf\n",sum/ans);
else
printf("Sorry!\n");
}
return 0;
}

  

HDU 2061 Treasure the new start, freshmen!的更多相关文章

  1. HDOJ(HDU) 2061 Treasure the new start, freshmen!(水题、)

    Problem Description background: A new semester comes , and the HDU also meets its 50th birthday. No ...

  2. hdu2061 Treasure the new start, freshmen!(暴力简单题)

    Treasure the new start, freshmen! Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/3276 ...

  3. HDU 3468 Treasure Hunting(BFS+网络流之最大流)

    题目地址:HDU 3468 这道题的关键在于能想到用网络流.然后还要想到用bfs来标记最短路中的点. 首先标记方法是,对每个集合点跑一次bfs,记录全部点到该点的最短距离.然后对于随意一对起始点来说, ...

  4. hdu 2061

    PS:  以为找个简单来恢复信心..结果碰到那么傻逼的题目... 题意:给出学分和成绩,算GPA...关键是注意换行....它要求的换行我觉得超级奇怪...除了第一个正常,其他的输入完之后先一个换行. ...

  5. HDU 3641 Treasure Hunting(阶乘素因子分解+二分)

    题目链接:pid=3641">传送门 题意: 求最小的 ( x! ) = 0 mod (a1^b1*a2^b2...an^bn) 分析: 首先吧a1~an进行素因子分解,然后统计下每一 ...

  6. hdu 3641 Treasure Hunting 强大的二分

    /** 大意:给定一组ai,bi . m = a1^b1 *a2^b2 * a3^ b3 * a4^b4*...*ai^bi 求最小的x!%m =0 思路: 将ai 质因子分解,若是x!%m=0 那么 ...

  7. 【网络流】 HDU 3468 Treasure Hunting

    题意: A-Z&&a-z 表示 集结点 从A点出发经过 最短步数 走到下一个集结点(A的下一个集结点为B ,Z的下一个集结点为a) 的路上遇到金子(*)则能够捡走(一个点仅仅能捡一次) ...

  8. HDU100题简要题解(2060~2069)

    这十题感觉是100题内相对较为麻烦的,有点搞我心态... HDU2060 Snooker 题目链接 Problem Description background: Philip likes to pl ...

  9. hdu2060-2062

    hdu 2060 斯诺克,读懂题意直接模拟 #include<stdio.h> int main(){ int N; ]; a[]=; ;i<=;i++){ a[i]=(-i)*i/ ...

随机推荐

  1. java rmi 入门实例

    java rmi 入门实例 (2009-06-16 16:07:55) 转载▼ 标签: java rmi 杂谈 分类: java-基础    java rmi即java远程接口调用,实现了2台虚拟机之 ...

  2. 【LNOI2014】LCA

    题面 题解 考察\(dep[\mathrm{LCA}(i, x)]\)的性质,发现它是\(i\)和\(x\)的链交的长度. 那么对每个\(i\)所在的链打一个区间加标记,询问时算一下\(x\)所在的链 ...

  3. bzoj 3232: 圈地游戏

    bzoj 3232: 圈地游戏 01分数规划,就是你要最大化\(\frac{\sum A}{\sum B}\),就二分这个值,\(\frac{\sum A}{\sum B} \geq mid\) \( ...

  4. [SYZOJ279]滑♂稽♂树

    主♂席♂树♂裸♂题 https://syzoj.com/problem/279 https://oj.changjun.com.cn/problem/detail/pid/2425 // It is ...

  5. 团队合作开发git冲突解决方案 Intellij IDEA

    https://blog.csdn.net/antony9118/article/details/52524873 注:因为我使用的是 idea 2018.1.2的版本,所以在操作上有些变化 将某些操 ...

  6. maven的pom文件报错: must be "pom" but is "jar"

    问题 Project build error: Invalid packaging for parent POM com.test:hello-parent:0.0.1-SNAPSHOT (E:\ec ...

  7. 大数据中Linux集群搭建与配置

    因测试需要,一共安装4台linux系统,在windows上用vm搭建. 对应4个IP为192.168.1.60.61.62.63,这里记录其中一台的搭建过程,其余的可以直接复制虚拟机,并修改相关配置即 ...

  8. C++模板的实现(模板函数和模板类,附带模板实现顺序表和链表代码)

    文章链接:https://blog.csdn.net/qq_38646470/article/details/80209469

  9. C++默认成员函数

    1.什么是面向对象? 概念:(Object Oriented Programming,缩写:OOP)是一种程序设计范型,同时也是一种程序开发的方法. 对象指的是类的实例,将对象作为程序的基本单元,将程 ...

  10. MySQL5.7(二)数据库的基本操作

    登录MySQL数据库 格式:mysql -u 用户名 -h 主机名或IP地址  -P 端口号 -p 密码