The mell hall
题目描述
In HUST,there are always manystudents go to the mell hall at the same time as soon as the bell rings.
Students have to queue up for a meal,and the queue is awalys long,So it takes much time.Suposse there
are N people in a queue,each personhas two characteristic value A and B(both of them are integers,read
input for more details),the i-thperson in the queue have to spend m(i) =
A1A2
:::Ai 1
Bi
minutes,Where
Ai
,Bi
is the i-th person.s valueA,B.Notice that if the order of the queue changes,the waiting time one
spend(that is,the value of m(i))maychanges too. Of course,every student want to reduce the time he
spend.
Apparently,it is impossible to makeeveryone satis
ed,in this problem,we only need to minimize the
waiting time of one who spend thelongest time in the queue,that is,minimize Max m(1),m(2),,,m(n).You
can change the order of the queue inanyway in order to complete this problem.
You are asked to output the originallocation in the queue of the person who will cost the longest time
under optimal solution. Uniquity isinsured by the given data.
输入
There are multiple test cases.
For each case, the
rst line contains one integerN(1N1000).
The second line containsNintegers Ai
.(0 < Ai <100000)
The third line containsNintergers Bi
(0< Bi <10).
(Bi <10< Ai*bi)
输出
You are asked to output the originallocation in the queue of the person who will cost the longest time
under optimal solution. Uniquity isinsured by the given data.
样例输入
3
15 20 25
1 3 2
样例输出
2
意思没看懂，听人解释后很简单。就是找出a[i]*b[i]的最大值的i值就行了！
#include<iostream>
using namespace std;
int main()
{
int n,i;
int a[100005],b[100005];
int m,t;
while(cin>>n)
{
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++)
cin>>b[i];
m=a[0]*b[0];
t=0;
for(i=1;i<n;i++)
{
if(a[i]*b[i]>m)
{
m=a[i]*b[i];
t=i;
}
}
cout<<t+1<<endl;
}
return 0;
}