Codeforces 1A (A. Theatre Square) Simple math problem

最新推荐文章于 2023-11-03 20:42:27 发布

转载最新推荐文章于 2023-11-03 20:42:27 发布 · 79 阅读

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原文链接：http://www.cnblogs.com/heat-man/archive/2013/03/10/2952666.html

博客围绕Berland首都剧院广场铺砖问题展开，广场为n×m米的矩形，要用a×a的正方形花岗岩石板铺设，要求计算最少所需石板数，输入为n、m、a三个正整数，还给出了AC代码及转载链接。

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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.

Sample test(s)

Input

6 6 4

Output

4



A simple math problem! But I was wrong 2 times at first.

Learned: 1 I learn how to use the function of "ceil".   eg:    ceil(4.3)=5;

         2 If the problem's data need to use the "__int64" type,you'd better insure  the relational data use the "__int64" type too,or you may wrong easily!


For this problem,I first submission was wrong because the data 'x' and 'y' are 'int', I should use the "__int64":

View Code

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 int main()
 5 {
 6   int m,n,a;
 7   int x,y;
 8   __int64 sum;
 9   while(cin>>m>>n>>a)
10   {
11       if(m%a==0)
12           x=m/a;
13       else
14           x=m/a+1;
15       if(n%a==0)
16           y=n/a;
17       else
18           y=n/a+1;
19       sum=x*y;
20       printf("%I64d\n",sum);
21 
22   }
23   return 0;
24 }

here is the AC code:

View Code

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 int main()
 5 {
 6   int m,n,a;
 7   __int64 x,y;
 8   __int64 sum;
 9   while(cin>>m>>n>>a)
10   {
11       if(m%a==0)
12           x=m/a;
13       else
14           x=m/a+1;
15       if(n%a==0)
16           y=n/a;
17       else
18           y=n/a+1;
19       sum=x*y;
20       printf("%I64d\n",sum);
21 
22   }
23   return 0;
24 }

转载于:https://www.cnblogs.com/heat-man/archive/2013/03/10/2952666.html