水

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, .., AN-1 representing the scores of other students in Bob's
class.

The last line contains M integers B1, B2, .., BM 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

2
4 3
5 5 5
4 4 3
6 5
5 5 4 5 3
1 3 2 2 1

Sample Output

4 4
2 4

Author: JIANG, Kai

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; const int INF=0x3f3f3f3f; int n,m;
int suma,sumb; int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
suma=sumb=0;
for(int i=0;i<n-1;i++)
{
int x;
scanf("%d",&x);
suma+=x;
}
for(int i=0;i<m;i++)
{
int x;
scanf("%d",&x);
sumb+=x;
}
int MIN=INF,MAX=-INF;
for(int i=0;i<=100;i++)
{
if((suma*n>(suma+i)*(n-1))&&(sumb*(m+1)<(sumb+i)*m))
{
MIN=min(MIN,i);
MAX=max(MAX,i);
}
}
printf("%d %d\n",MIN,MAX);
}
return 0;
}

ZOJ 3819 Average Score 水的更多相关文章

  1. [ACM] ZOJ 3819 Average Score (水题)

    Average Score Time Limit: 2 Seconds      Memory Limit: 65536 KB Bob is a freshman in Marjar Universi ...

  2. ZOJ 3819 Average Score(平均分)

    Description 题目描述 Bob is a freshman in Marjar University. He is clever and diligent. However, he is n ...

  3. zoj 3819 Average Score

    Average Score Time Limit: 2 Seconds      Memory Limit: 65536 KB Bob is a freshman in Marjar Universi ...

  4. ZOJ 3819 Average Score(数学 牡丹江游戏网站)

    主题链接:problemId=5373">http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373 Bob is ...

  5. ZOJ 2819 Average Score 牡丹江现场赛A题 水题/签到题

    ZOJ 2819 Average Score Time Limit: 2 Sec  Memory Limit: 60 MB 题目连接 http://acm.zju.edu.cn/onlinejudge ...

  6. 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 Universi ...

  7. 【解题报告】牡丹江现场赛之ABDIK ZOJ 3819 3820 3822 3827 3829

    那天在机房做的同步赛,比现场赛要慢了一小时开始,直播那边已经可以看到榜了,所以上来就知道A和I是水题,当时机房电脑出了点问题,就慢了好几分钟,12分钟才A掉第一题... A.Average Score ...

  8. UVa 1585 Score --- 水题

    题目大意:给出一个由O和X组成的串(长度为1-80),统计得分. 每个O的分数为目前连续出现的O的个数,例如,OOXXOXXOOO的得分为1+2+0+0+1+0+0+1+2+3 解题思路:用一个变量t ...

  9. ZOJ3819 ACM-ICPC 2014 亚洲区域赛的比赛现场牡丹江司A称号 Average Score 注册标题

    Average Score Time Limit: 2 Seconds      Memory Limit: 131072 KB Bob is a freshman in Marjar Univers ...

随机推荐

  1. 恩智浦Freescale Cortex-A9 迅为IMX6开发板平台初体验

    iTOP-i.MX6 开发板预装 Android4.4 系统,采用 9.7 寸(或者 7 寸或者 4.3 寸)IPS 屏 幕,至少 5 点以上触控,操作流畅,无论是高清视频.游戏等都会有上佳的表现,实 ...

  2. Swing实现个人简历

    源码: import java.awt.Container;import java.awt.FlowLayout;import java.awt.Font; import javax.swing.*; ...

  3. Android获取屏幕的大小与密度的代码

    Android项目开发中很多时候需要获取手机屏幕的宽高以及屏幕密度来进行动态布局,这里总结了三种获取屏幕大小和屏幕密度的方法 ? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...

  4. 诊断:AWR快照停止自动采集

    11.2.0.4数据库中,MMON进程,有时候由于一些莫名其妙的原因挂掉,接下来AWR的快照也就无法正常自动生成.MMON进程应该自动重启,却并没有自动被启动. 那么我们有可能是遇到了bug Bug ...

  5. 邮箱地址自动提示jQuery插件

    // mailAutoComplete.js v1.0 邮箱输入自动提示// 2010-06-18 v2.0 使用CSS class类代替CSS对象,同时增强代码可读性// 2010-06-18 v2 ...

  6. js获取当前位置

    <!DOCTYPE html><html><head><meta name="viewport" content="initia ...

  7. 高阶函数 map,reduce, filter的用法

    1. map 用法 def fun_C(x): """求平方""" return x ** 2 result = map(fun_C, my ...

  8. LeetCode(43)Multiply Strings

    题目 Given two numbers represented as strings, return multiplication of the numbers as a string. Note: ...

  9. 集训第五周动态规划 J题 括号匹配

    Description We give the following inductive definition of a “regular brackets” sequence: the empty s ...

  10. STM32F407 新建基于固件库的项目模板

    1.新建文件夹如图: 2.新建项目在USER文件夹中,选cpu如图: 若再弹出窗口, 直接点cancel 3.删了这俩文件夹: 4.复制文件到fwlib: src 存放的是固件库的.c 文件, inc ...