4000 CodeForces 1A Theatre Square 【数学】【精度】
2017-02-17 00:06
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Description
Theatre Square in the capital city of Berland has a rectangular shape with the sizen × 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 sizea × 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
Input
Output
题意:n*m的矩形区域放边长为a*a的矩形块,要求全部覆盖矩形区域的面积,且块不可分割,问最少需要多少块
思路:将面积问题转化为边长的问题。注意一下精度
附上几组数据:
1 1 1 1
1000000000 1000000000 1 1000000000000000000
1000000000 987654321 1 987654321000000000
AC代码:
Theatre Square in the capital city of Berland has a rectangular shape with the sizen × 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 sizea × 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
Input
6 6 4
Output
4
题意:n*m的矩形区域放边长为a*a的矩形块,要求全部覆盖矩形区域的面积,且块不可分割,问最少需要多少块
思路:将面积问题转化为边长的问题。注意一下精度
附上几组数据:
1 1 1 1
1000000000 1000000000 1 1000000000000000000
1000000000 987654321 1 987654321000000000
AC代码:
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> using namespace std; int main() { long long n, m, a; long long x, y; while(~scanf("%lld%lld%lld",&n, &m, &a)) { x = n / a; if(n%a) x++; y = m / a; if(m%a) y++; cout << x*y <<endl; } return 0; }
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