ZOJ 2819 Average Score
Time Limit: 2 Sec Memory Limit: 60 MB
题目连接
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5373
Description
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
HINT
题意给你两个班级，告诉你把bob扔到B班去，两个班级的平均分都会上升，然后问你，bob的分数上下限是多少
题解：
暴力枚举平均分，然后显然bob只要比第一个班级的平均分低，比第二个班级的平均分高就行了
代码：
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
*/
//**************************************************************************************
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
int main()
{
int t;
t=read();
while(t--)
{
int n,m;
n=read(),m=read();
int ans1=0,ans2=0;
for(int i=0;i<n-1;i++)
{
int x=read();
ans1+=x;
}
for(int i=0;i<m;i++)
{
int x=read();
ans2+=x;
}
int t1=0,t2=0;
while(t1*m<=ans2)
t1++;
while(t2*(n-1)<ans1)
t2++;
printf("%d %d\n",t1,t2-1);
}
return 0;
}