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 isGPA = (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. InputThe 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 legalOutputOutput 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 Input23Algorithm 3 97DataStruct 3 90softwareProject 4 852Database 4 59English 4 81Sample Output90.10sorry
題解：這道水題很簡(jiǎn)單，值得注意的是每?jī)山M數(shù)據(jù)之間要有兩個(gè)空格。最后一個(gè)數(shù)據(jù)沒(méi)有空格。我的處理辦法是在最后加了一個(gè) if（n） 來(lái)判斷。
我的代碼：
#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<iomanip>#include<string.h>using namespace std;int main( ){ int n, num; cin>> n; while (n--) { cin >> num; char s[16000]; memset (s, '0', sizeof(s)); double a[1000] = {0}, b[1000] = {0}, r = 0, c = 0; int i, flag=0; for (i = 1; i <= num; i++) cin >> s >> a[i] >> b[i]; for (i = 1; i <= num; i++) { r = a[i] * b[i] + r; c = c + a[i]; } r = r / c; for (i = 1; i <= num; i++) { if(b[i] < 60) flag = 1; } if (flag == 1) cout << "Sorry!" << endl; else printf("%.2lf/n", r); if (n) cout <<endl; } return 0;}
