zoj 3819 Average Score 【签到题】

最新推荐文章于 2019-04-11 08:40:00 发布

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最新推荐文章于 2019-04-11 08:40:00 发布

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本文介绍了一个数学问题，即如何计算Bob在两个班级间转移时，为了使两个班级的平均分都提高，他的成绩可能的最小值和最大值。通过已知条件分析，给出了具体的算法实现思路及AC代码。

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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) 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 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

题意：Bob班里有N个人（加上自己），邻班有M个人。若Bob不在自己的班里，而在邻班里面，两班的平均分都会增加。现在已经给出Bob班里N-1个人的分数以及邻班M个人的分数，问你Bob分数的最小值和最大值。

设Bob分数为x，自己班里除了他之外总分数为sum1，邻班总分数为sum2。
显然有sum1/ (N-1) >= x >= sum2 / M + 1。对于sum1 / (N-1)要分能否整除讨论一下。

AC代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#define LL long long
#define INF 0x3f3f3f3f
#define MAXN 1000
#define MAXM 100000
using namespace std;
int a[60], b[60];
int main()
{
    int t;
    int N, M;
    int sum1, sum2;
    int p1, p2;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &N, &M);
        sum1 = sum2 = 0;
        for(int i = 0; i < N-1; i++)
        {
            scanf("%d", &a[i]);
            sum1 += a[i];
        }
        if(sum1 % (N-1) == 0)
            p1 = sum1 / (N-1) - 1;
        else
            p1 = sum1 / (N-1);
        for(int i = 0; i < M; i++)
        {
            scanf("%d", &b[i]);
            sum2 += b[i];
        }
        p2 = sum2 / M;
        printf("%d %d\n", p2+1, p1);
    }
    return 0;
}