HDU - 4802 GPA

Description

In college, a student may take several courses. for each course i, he earns a certain credit (c _i), and a mark ranging from A to F, which is comparable to a score (s _i), according to the following conversion table 

The GPA is the weighted average score of all courses one student may take, if we treat the credit as the weight. In other words, 

An additional treatment is taken for special cases. Some courses are based on “Pass/Not pass” policy, where stude nts earn a mark “P” for “Pass” and a mark “N” for “Not pass”. Such courses are not supposed to be considered in computation. These special courses must be ignored for computing the correct GPA. 
Specially, if a student’s credit in GPA computation is 0, his/her GPA will be “0.00”.

 

Input

There are several test cases, please process till EOF. 
Each test case starts with a line containing one integer N (1 <= N <= 1000), the number of courses. Then follows N lines, each consisting the credit and the mark of one course. Credit is a positive integer and less than 10.

 

Output

For each test case, print the GPA (rounded to two decimal places) as the answer.

 

Sample Input

 52 B3 D-2 P1 F3 A22 P2 N64 A3 A3 A4 A3 A3 A 

 

Sample Output

 2.330.004.00 

题意：比较水的题目

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;map<string, double> mp;int n;int main() {int m;string str;mp["A"] = 4.0, mp["A-"] = 3.7, mp["B+"] = 3.3;mp["B"] = 3.0, mp["B-"] = 2.7, mp["C+"] = 2.3;mp["C"] = 2.0, mp["C-"] = 1.7, mp["D"] = 1.3;mp["D-"] = 1.0, mp["F"] = 0.0;while (scanf("%d", &n) != EOF) {double ans = 0.00;int sum = 0;for (int i = 0; i < n; i++) {cin >> m >> str;if (str == "P" || str == "N")continue;sum += m;ans += (double) m * mp[str];}if (sum == 0)printf("0.00\n");else printf("%.2f\n", 1.0*ans/sum);}return 0;}