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 (ci), and a mark ranging from A to F, which is comparable to a score (si), 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”.
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.
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[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
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"]=;
H["N"]=;
H["P"]=;
while(scanf("%d",&n)!=EOF)
{
double num=;
double kiss=;
for(int i=;i<=n;i++)
{
int tmp=read();
string s;
cin>>s;
if(s=="N"||s=="P")
tmp=;
kiss+=H[s]*tmp;
num+=tmp;
}
if(num==)
cout<<"0.00"<<endl;
else
printf("%.2lf\n",kiss/num);
}
}