Problem--1A--Codeforces--TheatreSquare
2017-01-17 16:26
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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
input
6 6 4
output
4
Codeforces第一题–向上取整问题
题意大致为:已知面积为m*n的地方,用大小为a*a的砖来铺满,但是砖不可以弄碎,求所用的最少砖数(分别对长和宽向上取整即可).
自己的代码不简洁,看了大神们的代码,才知道缺陷在哪里,顺便记录下来,弥补不足.
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
input
6 6 4
output
4
Codeforces第一题–向上取整问题
题意大致为:已知面积为m*n的地方,用大小为a*a的砖来铺满,但是砖不可以弄碎,求所用的最少砖数(分别对长和宽向上取整即可).
自己的代码不简洁,看了大神们的代码,才知道缺陷在哪里,顺便记录下来,弥补不足.
#include<stdio.h> int main() { int m,n,a; scanf("%d%d%d",&m,&n,&a); printf("%I64d\n",(m/a+(m%a>0))*(n/a+(n%a>0))*1ll); return 0; }
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