杭电acm2061Treasure the new start, freshmen!(珍惜新的开始,新生!GPA)
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Treasure the new start, freshmen!
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12231 Accepted Submission(s): 3780
Problem Description
background:
A new semester comes , and the HDU also meets its 50th birthday. No matter what's your major, the only thing I want to tell you is:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school.
"No pain, No Gain", HDU also has scholarship, who can win it? That's mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.
If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA is
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
If there is a 0 <= Si < 60, The GPA is always not existed.
A new semester comes , and the HDU also meets its 50th birthday. No matter what's your major, the only thing I want to tell you is:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school.
"No pain, No Gain", HDU also has scholarship, who can win it? That's mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.
If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA is
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
If there is a 0 <= Si < 60, The GPA is always not existed.
Input
The first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).
Notice: There is no blank in the Course Name. All the Inputs are legal
Notice: There is no blank in the Course Name. All the Inputs are legal
Output
Output the GPA of each case as discribed above, if the GPA is not existed, ouput:"Sorry!", else just output the GPA value which is rounded to the 2 digits after the decimal point. There is a blank line between two test cases.
Sample Input
23Algorithm 3 97DataStruct 3 90softwareProject 4 852Database 4 59English 4 81
Sample Output
90.10Sorry!
Author
zl
注:此题为:杭电OJ 2061Treasure the new start, freshmen!(珍惜新的开始,新生!GPA)
题意:
就是算一个平均分(在其中出现过60以下的,输出“Sorry!”),每组测试数据之间空一行,最后一个不空。
思路:
如果出现60以下的,提早做标记
求总值数 fz=SUM(学分×单科成绩),总分数 fm=SUM(学分)
得出平均分 fz/fm;
注意点: 在最后一个案例不空行!
以下为AC代码:
#include<iostream>#include<cstdio> using namespace std;int main(){ int N; cin>>N; while(N--) { int K; cin>>K; char ch[200][200]; int i,j,bz=0; double GPA,c[200],s[200],fz=0,fm=0; for(j=0,i=0;i<K;++i,++j) { getchar(); scanf("%s %lf%lf",ch[j],&c[i],&s[i]); } for(i=0;i<K;++i) if(s[i]<60) { bz=1; break; } if(bz) cout<<"Sorry!"<<endl; else { for(i=0;i<K;++i) { fz+=c[i]*s[i]; fm+=c[i]; } GPA=fz/fm; printf("%.2lf\n",GPA); } if(N!=0) cout<<endl; } return 0;}
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