Problem : Clockwise Bricks Breaker
Clockwise Bricks Breaker is a game where there is a ball which is used to break the block of bricks. Brick's Blocks are always in a form of a square matrix( e.g 2X2, 4X4). The ball has some characteristics like for breaking each brick from the block. It consumes 1 unit of energy and the whole setup is in a manner that the ball will break bricks always - in clock wise direction, in a ring fashion and start from top-left. Take an example of 4X4 brick block.
Figure 1.
Figure 2.
Figure 3.
Your Task is to calculate total points (points of a block should be the positional value of that block as mentioned in the picture: You have to add points of all the broken bricks) gained by a player, where the energy of ball and size of block will be given.
Input Format:
First line of input should be the number of test cases T
Next T lines contain two variables N and E delimited by a whitespace where,
First line of input should be the number of test cases T
Next T lines contain two variables N and E delimited by a whitespace where,
- N =size of square block (i.e. block will be of NXN bricks)
- E =Energy of ball
Output Format:
Output will be T lines containing the integer which is total number points gained by that energy.
Output will be T lines containing the integer which is total number points gained by that energy.
OR
Print "Invalid Input" if constraint fails
Print "Invalid Input" if constraint fails
Constraints:
- 0< T<=10
- 0< N <=200
- 0<= E <=N*N
Examp
| EXAMPLE NUMBER | SAMPLE INPUT | SAMPLE OUTPUT | EXPLAINATION |
|---|---|---|---|
| 1 | 2 4 5 1 50 | 36 Invalid Input | For first test case as with Energy 5, the ball will break 5 bricks having points 1,4,16,13,2 So 1+4+16+13+2=36. Second test case is Invalid Input because constraint E<=N*N not satisfied. |
| 2 | 2 5 0 9 -3 | 0 Invalid Input | For first test case it is 0 as energy is zero. For second test case it is Invalid Input because constraint E >= 0 not satisfied. |
SOLUTION :
#include<iostream>
using namespace std;
int main()
{
int N,P;
cout<<"Enter N & Power :";
cin>>N>>P;
if(P>N*N)
P=N*N;
int wall[N][N],temp=0;
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
temp++,wall[i][j]=temp;
int sum=0,
i[4]={0,0,N-1,N-1},
j[4]={0,N-1,N-1,0};
temp=0;
while(P>0)
{
if(temp%4==0)
{
if(wall[i[0]][j[0]]==0)
{
i[0]+=1;
j[0]=i[0];
sum+=wall[i[0]][j[0]];
wall[i[0]][j[0]]=0;
j[0]+=1;
}
else
{
sum+=wall[i[0]][j[0]];
wall[i[0]][j[0]]=0;
j[0]+=1;
}
}
if(temp%4==1)
{
if(wall[i[1]][j[1]]==0)
{
j[1]-=1;
i[1]=i[0];
sum+=wall[i[1]][j[1]];
wall[i[1]][j[1]]=0;
i[1]+=1;
}
else
{
sum+=wall[i[1]][j[1]];
wall[i[1]][j[1]]=0;
i[1]+=1;
}
}
if(temp%4==2)
{
if(wall[i[2]][j[2]]==0)
{
i[2]-=1;
j[2]=i[2];
sum+=wall[i[2]][j[2]];
wall[i[2]][j[2]]=0;
j[2]-=1;
}
else
{
sum+=wall[i[2]][j[2]];
wall[i[2]][j[2]]=0;
j[2]-=1;
}
}
if(temp%4==3)
{
if(wall[i[3]][j[3]]==0)
{
j[3]+=1;
i[3]=i[2];
sum+=wall[i[3]][j[3]];
wall[i[3]][j[3]]=0;
i[3]-=1;
}
else
{
sum+=wall[i[3]][j[3]];
wall[i[3]][j[3]]=0;
i[3]-=1;
}
}
temp+=1;
P--;
}
cout<<"sum="<<sum;
}
