觉得浮夸了四年,漠然发现原来是浮躁了四年!

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

A. Theatre Square
time limit per test
2 seconds
memory limit per test
64 megabytes
input
standard input
output
standard 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.

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":
复制代码
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 }
复制代码


here is the AC code:

复制代码
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 }
复制代码

 

posted @   heat nan  阅读(2736)  评论(0编辑  收藏  举报
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