【codeforces 1A】Theatre Square——数学,水
2016-04-25 15:10
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题目:
A. Theatre Square
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Theatre Square in the capital city of Berland has a rectangular shape with the size n × mmeters. 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
input
output
描述:给定一个n*m的矩形,要求用a*a的正方形块实现全覆盖,可以超出矩形的面积
题解:看矩形的两条边分别需要多少a才能全覆盖。
代码:
A. Theatre Square
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Theatre Square in the capital city of Berland has a rectangular shape with the size n × mmeters. 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
input
6 6 4
output
4
描述:给定一个n*m的矩形,要求用a*a的正方形块实现全覆盖,可以超出矩形的面积
题解:看矩形的两条边分别需要多少a才能全覆盖。
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { int n,m; scanf("%d%d",&n,&m); int a; scanf("%d",&a); __int64 l,r; l = (n % a == 0) ? n / a : n / a + 1; r = (m % a == 0) ? m / a : m / a + 1; printf("%I64d\n",l * r); return 0; }
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