HDU 2399 GPA

GPA

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2502    Accepted Submission(s): 1469

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

解题思路：A，B，C，D，F各对应的分数为4，3，2，1，0，给出一行数据，每个字母用空格隔开，只能包含A，B，C，D，F和“ ”，若还有其他字符，则输出 ”Unknown letter grade in input“；若满足条件，则计算他们的平均值。

AC代码：

#include <iostream>#include <cstdio>#include <cstring> using namespace std;char s[500]; int main(){ //freopen("in.txt", "r", stdin); while(gets(s)){ int len = strlen(s); int cnt = 0; int k = 0; int flag = 1; for(int i=0; i<len; i++){ if(s[i] == 'A') { cnt += 4; k++; } else if(s[i] == 'B') { cnt += 3; k++; } else if(s[i] == 'C') { cnt += 2; k++; } else if(s[i] == 'D') { cnt += 1; k++; } else if(s[i] == 'F') { cnt += 0; k++; } else if(s[i] == ' ') continue; else{ flag = 0; break; } } if(!flag) printf("Unknown letter grade in input\n"); else printf("%.2f\n", (float)cnt/k); } return 0; }