hdu 4802 GPA 水题

GPA

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=4802

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

5
2 B
3 D-
2 P
1 F
3 A
2
2 P
2 N
6
4 A
3 A
3 A
4 A
3 A
3 A

Sample Output

2.33 0.00 4.00

 

HINT

 For the first test case: GPA =(3.0 * 2 + 1.0 * 3 + 0.0 * 1 + 4.0 * 3)/(2 + 3 + 1 + 3) = 2.33 For the second test case: because credit in GPA computation is 0(P/N in additional treatment), so his/her GPA is “0.00”.

题意

算GPA！

题解：

签到题

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 200001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

map<string,double> H;
int main()
{
    //test;
    int n;
    H["A"]=4.0;
    H["A-"]=3.7;
    H["B+"]=3.3;
    H["B"]=3.0;
    H["B-"]=2.7;
    H["C+"]=2.3;
    H["C"]=2.0;
    H["C-"]=1.7;
    H["D"]=1.3;
    H["D-"]=1.0;
    H["F"]=0;
    H["N"]=0;
    H["P"]=0;
    while(scanf("%d",&n)!=EOF)
    {
        double num=0;
        double kiss=0;
        for(int i=1;i<=n;i++)
        {  
            int tmp=read();
            string s;
            cin>>s;
            if(s=="N"||s=="P")
                tmp=0;
            kiss+=H[s]*tmp;
            num+=tmp;
        }
        if(num==0)
            cout<<"0.00"<<endl;
        else
            printf("%.2lf
",kiss/num);
    }
}