1778 credit points and

 

Description

Credit points and are the main basis for the final ranking of Beijing University of Aeronautics and Astronautics. Their calculation rules are to multiply the credits of each required course by the scores obtained from this course. The formula for calculating the scores of each course is as follows:

If the score is greater than or equal to 90, the score is recorded as 1.5.

If the score is greater than or equal to 80 and less than 90, the score is 1.3.

If the score is greater than or equal to 70 and less than 80, the score is 1.1.

If the score is greater than or equal to 60 and less than 70, the score is 1.0.

If the score is less than 60, the score is 0.0.

The excellent, good, medium, pass, and fail scores of sports compulsory (and sports extracurricular) correspond to 1.5, 1.3, 1.1, 1.0, and 0 respectively.

Because the score is given by the teachers of different subjects, the final time for each subject is usually different. Some time sb. knows the score of several subjects, so please help calculate his/her average score and what is it?

To simplify the problem, physical education classes are not considered here.

By the way, I forgot to say that the average point and two decimal places are retained ).

Input

Multiple groups of data, the first behavior data group T

For each group of data, the first is a number N, indicating sb. I already know the scores of N subjects, followed by N pairs of real AI, Bi for the n subjects of credits and Sb..

Output

For each group of data, a row is output as a real number, indicating the average performance point and Sb.

Solution: This is a simulated question. Pay attention to the interval to make a judgment. Add the corresponding score points.

 

#include <iostream>#include <cstdio>using namespace std;int main(){int number;double sum,c;int a,b;cin>>number;for(int i=0;i<number;i++){int temp;sum=0;cin>>temp;for(int j=0;j<temp;j++){cin>>a>>b;if(b>=90)c=1.5;else{if(b>=80)c=1.3;else{if(b>=70)c=1.1;else{if(b>=60)c=1.0;elsec=0;}}}          sum+=a*c;}printf("%.2lf\n",sum);}return 0;}