1408201651-hd-GPA.cpp

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最新推荐文章于 2025-03-02 23:22:43 发布

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本文介绍了一个简单的GPA计算器程序，能够处理一系列大写字母形式的成绩，并计算出平均GPA。如果输入包含不属于标准成绩等级的字母，则会输出错误信息。

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GPA

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3341 Accepted Submission(s): 1354

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

题目大意

给你一系列大写字母代表学生的成绩。A=4，B=3，C=2，D=1，F=0。输出该学生的平均分数，若出现不属于{'A','B','C','D','F'}的大写字母，则输出Unknown letter grade in input。

错误原因

一开始以为除了那五个大写字母之外，会出现的只有'E'，这才导致了错误。

解题思路

因为不确定给出你的大写字母的次数，所以你只能用gets()输入一串字符串。然后在对' '(空格)另外考虑。

代码

<span style="font-size:18px;">#include<stdio.h>
#include<string.h>
char s[1000];
int in(char a)
{
    if(a=='A')
        return 4;
    else if(a=='B')
        return 3;
    else if(a=='C')
        return 2;
    else if(a=='D')
        return 1;
    else if(a=='F')
        return 0;
    else if(a==' ')
        return 5;
    else
        return -1;
    //输入的只有空格和大写字母(A、B、C、D、E、F、G、H...) 
}
int main()
{
    int i,j,k,l,sum;
    int len;
    while(gets(s)!=NULL)
    {
        len=strlen(s);
        sum=0;
        k=0;
        l=0;
        for(i=0;i<len;i++)
        {
            j=in(s[i]);
            if(j>-1&&j<5)
            {
                sum+=j;
                l++;
            }
            else if(j==-1)
            {
                k=1;
                break;
            }
        }
        if(k==1)
            printf("Unknown letter grade in input\n");
        else
            printf("%.2lf\n",sum/(l*1.0));
    }
    return 0;
}
</span>