A. Theatre Square（codefore）

A. Theatre Square
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the occasion of the city’s anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size a × a.

What is the least number of flagstones needed to pave the Square? It’s allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It’s not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

Input
The input contains three positive integer numbers in the first line: n,  m and a (1 ≤  n, m, a ≤ 109).

Output
Write the needed number of flagstones.

Examples
inputCopy
6 6 4
outputCopy
4
The question ： It's simple ,
AC：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
int main()
{
    
	int n,m,a;
	cin>>n>>m>>a;

	ll x=ceil(n*1.0/a)*ceil(m*1.0/a);
	//ll y=ceil(m*1.0/a);
	cout<<x<<endl;
    return 0;
}

ceil,,math function , Rounding up

Error code ： Although it looks complicated , I just put the situation out for discussion and output , Even if I use long long, Still 1000000000 1000000000 1 In this case ,codeforce say Signed integer overflow . I don't know what I mean , But the code above will not , When I made this mistake, I still thought that the problem of high probability should be multiplied with high precision , Ha ha ha ,！！！！

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<cstdio>
#define ll long long
using namespace std;
int main()
{
    
	int n,m,a;
	cin>>n>>m>>a;
	ll sum;
	if(n%a==0)
	{
    
		if(m%a==0)
		{
    
			//ll sum;
			sum=n/a*(m/a);// Signed integer overflow 
			cout<<sum<<endl;
		}
		
		else
		{
    
			sum=n/a*(m/a+1);
				cout<<sum<<endl;
		}
	
	 } 
	 else
	 {
    
	 	if(m%a==0)
	 	{
    
	 		sum=m/a*(n/a+1);
	 		cout<<sum<<endl;
		 }
		 else
		 {
    
		 	sum=(n/a+1)*(m/a+1);
		 	cout<<sum<<endl;
		 }
		 
	 }
    return 0;
}