Here to remember:
– Theatre square is rectangular shape n X m meters
– Each flagstone is of the size a X a
What is the least number of flagstones needed to pave the Square ?
Condition:
-Cover of more than Theatre area will be allowed but Theatre Square need to covered.
-Flagstones will not be broken.It must be a X a
-The sides of flagstones should be parallel to the sides of the square
So algorithm/Hints:
1. Input n,m,a
2. rowside=n/a;
3. coloumnside=m/a;
4. We don’t need point value so x=ceil(rowside),y=ceil(coloumnside)
5. pave the whole square with x*y;
why n/a or m/a : we need to find out this relation using some brain storming to apply the equation in code and we need some mathematics tricks for this
code:
#include<bits/stdc++.h> using namespace std; int main() { double n,m,a; while(scanf("%lf%lf%lf",&n,&m,&a)==3) { printf("%.0lf\n",ceil(n/a)*ceil(m/a)); } return 0; }