Zoj 3819 average score (mathematical problem Mudanjiang)

Zju math zoj Mudanjiang

Question link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 5373

Bob is a freshman in marjar University. He is clever and diligent. However, he is not good at math, especially in mathematical analysis.

After a mid-term exam, Bob was anxious about his grade. He went to the specified sor asking about the result of the exam. The specified sor said:

"Too bad! You made me so disappointed ."

"Hummm... I am giving lessons to two classes. If you were in the other class, the average scores of both classes will increase ."

Now, you are given the scores of all students in the two classes, Except t for the Bob's. please calculate the possible range of Bob's score. all scores shall be integers within [0,100].

Input

There are multiple test cases. The first line of input contains an integerTIndicating the number of test cases. For each test case:

The first line contains two integersN(2 <=N<= 50) andM(1 <=M<= 50) indicating the number of students in Bob's class and the number of students in the other class respectively.

The next line containsN-1 IntegersA1,A2,..,An-1Representing the scores of other students in Bob's class.

The last line containsMIntegersB1,B2,..,BMRepresenting the scores of students in the other class.

Output

For each test case, output two integers representing the minimal possible score and the maximal possible score of Bob.

It is guaranteed that the solution always exists.

Sample Input

24 35 5 54 4 36 55 5 4 5 31 3 2 2 1

Sample output

4 42 4

Author: Jiang, Kai

PS: 2014 ACM/ICPC Asian contest Mudanjiang (first stop) ___ sign-in question!
The Code is as follows:

#include <cstdio>#include <cmath>int main(){    int t;    int n, m;    int minn, maxx;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        double sum1=0, sum2=0;        int a;        for(int i = 0; i < n-1; i++)        {            scanf("%d",&a);            sum1+=a;        }        for(int i = 0; i < m; i++)        {            scanf("%d",&a);            sum2+=a;        }        double avg1=(sum1/(n-1.0));        double avg2=sum2/(m*1.0);        if((int)avg1==avg1)            maxx=(int)avg1-1;        else            maxx=(int)avg1;        minn=(int)avg2+1;        printf("%d %d\n",minn,maxx);    }    return 0;}

Zoj 3819 average score (mathematical problem Mudanjiang)