KOPC12A  K12  Building Construction
Given N buildings of height h1,h2,h3...hn, the objective is to make every building has equal height. This can be done by removing bricks from a building or adding some bricks to a building.Removing a brick or adding a brick is done at certain cost which will be given along with the heights of the buildings.Find the minimal cost at which you can make the buildings look beautiful by reconstructing the buildings such that the N buildings satisfy
h1=h2=h3=..=hn=k ( k can be any number).
For convenience, all buildings are considered to be vertical piles of bricks, which are of same dimensions.
Input
The first line of input contains an integer T which denotes number of test cases .This will be followed by 3*T lines , 3 lines per test case. The first line of each test case contains an integer n and the second line contains n integers which denotes the heights of the buildings [h1,h2,h3....hn] and the third line contains n integers [c1,c2,c3...cn] which denotes the cost of adding or removing one unit of brick from the corresponding building.
T<=15;n<=10000;0<=Hi<=10000;0<=Ci<=10000;
Output
The output must contain T lines each line corresponding to a testcase.
Example
Input:
1 3 1 2 3 10 100 1000 Output:
120
hide comments
LeppyR64:
20120403 04:13:13
K=3 All n buildings must be the same height. Height of zero doesn't decrease n. removing from one building is an action on that building. Adding it to another is another action on the other building. 10 to add a brick to the frst building twice is 20. 100 to add a brick to second building is 120. 

Dejan Tomic:
20120402 22:56:11
Wait, I think I understand. Does the sum of all heights has to be the same at the end? If the answer is NO then I understand everything now... 

Dejan Tomic:
20120402 21:37:02
I just do not understand the test case. Why is the solution 120?

Added by:  Radhakrishnan Venkataramani 
Date:  20120131 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  Own 