• GPA


    原题:

    GPA

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3645 Accepted Submission(s): 1489
     
    Problem Description
    Each course grade is one of the following five letters: A, B, C, D, and F. (Note that there is no grade E.) The grade A indicates superior achievement , whereas F stands for failure. In order to calculate the GPA, the letter grades A, B, C, D, and F are assigned the following grade points, respectively: 4, 3, 2, 1, and 0.
     
    Input
    The input file will contain data for one or more test cases, one test case per line. On each line there will be one or more upper case letters, separated by blank spaces.
     
    Output
    Each line of input will result in exactly one line of output. If all upper case letters on a particular line of input came from the set {A, B, C, D, F} then the output will consist of the GPA, displayed with a precision of two decimal places. Otherwise, the message "Unknown letter grade in input" will be printed.
     
    Sample Input
    A B C D F
    B F F C C A
    D C E F
     
    Sample Output
    2.00
    1.83
    Unknown letter grade in input
     
    Author
    2006Rocky Mountain Warmup
     
    Source
    HDU “Valentines Day” Open Programming Contest 2009-02-14
     
    Recommend
    lcy

    源代码:

     1 #include <iostream>
     2 #include <iomanip>
     3 #include <string>
     4 using namespace std;
     5 
     6 int main()    {
     7     double result;
     8     string bbb;
     9     int sum = 0, count = 0;
    10     char aaa;
    11     while (getline(cin, bbb))    {
    12         bool judge = true;
    13         for (int i = 0; i < bbb.length(); i++)    {
    14             aaa = bbb[i];
    15             switch (aaa)    {
    16                 case 'A':{
    17                     sum += 4;
    18                     count++;
    19                     break;
    20                 }
    21                 case 'B':{
    22                     sum += 3;
    23                     count++;
    24                     break;
    25                 }
    26                 case 'C':{
    27                     sum += 2;
    28                     count++; 
    29                     break;
    30                 }
    31                 case 'D':{
    32                     sum += 1;
    33                     count++;
    34                     break;
    35                 }
    36                 case 'F':{
    37                     count++;
    38                     break;
    39                 }
    40                 case ' ':{
    41                     break;
    42                 }
    43                 default:{
    44                     judge = false;
    45                     break;
    46                 }
    47             } 
    48         } 
    49         result = sum * 1.0 / count; // sum, count都是int型,一定记住在两个int型要转换成double型时,要乘1.0 
    50         if (judge)    cout << fixed << setprecision(2) << result << endl; // setpresicion的单词要拼写对 
    51         else    cout << "Unknown letter grade in input" << endl; 
    52         sum = 0;    count = 0; //记得初始化为0 
    53     }
    54     return 0;
    55 } 
  • 相关阅读:
    homebrew 安装 mpv
    Spring JdbcTemplate 两种方法的区别
    git .gitignore失效的解决办法
    git 分支修改bug应用场景
    url编码实践
    escape encodeuri encodeURIComponent 区别
    mysql命令gruop by报错this is incompatible with sql_mode=only_full_group_by
    服务器病毒问题解决- 阿里云 挖矿病毒,Circle_MI.png
    trim和replace的陷阱实践
    mysql 5.7.15 union order by 子查询排序不生效
  • 原文地址:https://www.cnblogs.com/QingHuan/p/4272634.html
Copyright © 2020-2023  润新知