ZOJ 3819 Average Score
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3819
题面：
Average Score Time Limit: 2 Seconds Memory Limit: 65536 KBBob 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
题意：
该学生在A班会拉低该班成绩，在B班会提高该班成绩，求该学生的成绩能取到的最大值和最小值。因为是整数有点特别，注意下整除的情况就好。
代码1：
#include <iostream>
using namespace std;
int main()
{
int t,n,m,sum1,sum2,tmp,minn,maxx;
cin>>t;
while(t--)
{
sum1=sum2=0;
cin>>n>>m;
for(int i=1;i<n;i++)
{
cin>>tmp;
sum1+=tmp;
}
for(int i=0;i<m;i++)
{
cin>>tmp;
sum2+=tmp;
}
if(sum2%m==0)minn=sum2/m+1;
else minn=(1.0*sum2/m)+1;
if(sum1%(n-1)==0)maxx=sum1/(n-1)-1;
else maxx=sum1/(n-1);
cout<<minn<<" "<<maxx<<endl;
}
return 0;
}
代码2：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <cmath>
#include <cstdlib>
#include <set>
#include <algorithm>
#define LL long long
using namespace std;
LL a,b,s;
int main()
{
int total1,n,m,total2,minn,maxx,tmp,t;
cin>>t;
while(t--)
{
cin>>n>>m;
total1=total2=0;
for(int i=1;i<n;i++)
{
cin>>tmp;
total1+=tmp;
}
for(int i=1;i<=m;i++)
{
cin>>tmp;
total2+=tmp;
}
for(int i=0;i<=100;i++)
{
if(i*(n-1)>=total1)
{
maxx=i-1;
break;
}
}
for(int i=100;i>=0;i--)
{
if(i*m<=total2)
{
minn=i+1;
break;
}
}
cout<<minn<<" "<<maxx<<endl;
}
return 0;
}