2014ACM/ICPC亚洲区域赛牡丹江站现场赛-A （ ZOJ 3819 ） Average Score

Average Score

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 professor asking about the result of the exam. The professor 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 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 integer T indicating the number of test cases. For each test case:

The first line contains two integers N (2 <= N <= 50) and M (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 contains N - 1 integers A₁, A₂, .., A_N-1 representing the scores of other students in Bob's
class.

The last line contains M integers B₁, B₂, .., B_M representing 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

Sample Output

4 4

2 4

Author: JIANG, Kai

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

题目链接：Average Score

解题思路：牡丹江站的签到题！！

！好恶心，当时在现场，好像快一个小时的时候才出题。太菜了。。。開始还没想到简单的方法，想直接暴力枚举，后来又想了想，直接计算出A班的n-1个人的平均成绩，然后取下界。这就是分数的最大值。再求B班的m个人的平均成绩，取上界。这就是分数的下界。

AC代码：

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <time.h>

using namespace std;

#define INF 0x7fffffff

int main()

{

    #ifdef sxk

        freopen("in.txt","r",stdin);

    #endif

    int T, n, m, sum, c;

    scanf("%d",&T);

    while(T--)

    {

        sum = 0;

        scanf("%d%d",&n,&m);

        for(int i=0; i<n-1; i++){

            scanf("%d", &c);

            sum += c;

        }

        double ans1 = 1.0*sum/(n-1);

        if(floor(ans1) == ans1)  ans1 --;

        else ans1 = floor(ans1);       //取下界

        sum = 0;

        for(int i=0; i<m; i++){

            scanf("%d", &c);

            sum += c;

        }

        double ans2 = 1.0*sum/m;

        if(floor(ans2) == ans2)  ans2 ++;   //取上界

        else ans2 = ceil(ans2);

        cout<<ans2<<" "<<ans1<<endl;

    }

    return 0;

}