HDU 1.2.5 GPA(ACM step)

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 FB F F C C AD C E F

 

Sample Output

2.001.83Unknown letter grade in input

 

Author

2006Rocky Mountain Warmup

 

Source

HDU “Valentines Day” Open Programming Contest 2009-02-14

 

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lcy

/*其实做个题太水了，不过就是在输入输出上我一时没转过弯儿连续输入一行字母，一旦回车就结束，打印出平均分 原来试了试不开数组直接用变量保存总分，没成功， 原因我忘了，所以只好开了个数组存每次的成绩如果谁可以不开数组，能直接回复我一下吗 */#include <stdio.h>#include <string.h>void GPA(char *grade, int cnt);int main(void){char ch, grade[1000] = {0};int cnt = 0;int i;while(scanf("%c", &ch)!=EOF){if(ch >= 'A' && ch <= 'z'){grade[cnt] = ch; // 记录分数 cnt ++;}if(ch == '\n'){GPA(grade, cnt);memset(grade, 0, cnt*sizeof(char));cnt = 0;}}return 0;}void GPA(char *grade, int cnt){int i, Unknown = 0;double sum = 0;for(i = 0; i < cnt; i++){switch(grade[i]){case 'A': sum += 4; continue;case 'B': sum += 3; continue;case 'C': sum += 2; continue;case 'D': sum += 1; continue;case 'F': continue;default : Unknown = 1; break;}}if(Unknown == 0)printf("%.2lf\n", sum / cnt);elseprintf("Unknown letter grade in input\n");}