CodeForces 1A Theatre Square
2016-05-15 15:56
447 查看
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 × 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
output
题意:给出一个长方形的长和宽,以及一个正方形的边长,问需要多少个正方形能把这个长方形全部覆盖。
水题。。。。。。。
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 × 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
题意:给出一个长方形的长和宽,以及一个正方形的边长,问需要多少个正方形能把这个长方形全部覆盖。
水题。。。。。。。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { __int64 a,b,c; while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF) { __int64 sum1=0,sum2=0; if(a%c!=0) { sum1=a/c+1; } else sum1=a/c; if(b%c!=0) { sum2=b/c+1; } else sum2=b/c; printf("%I64d\n",sum1*sum2); } return 0; }
相关文章推荐
- java在多网口的机器上的端口监听
- [系统开发] 一个基于Django和PureCSS的内容管理系统
- 一万小时定律的数学解释
- 【hibernate进阶】如何在myeclipse中添加jar包
- 菜单条、菜单、菜单项
- 模拟实现通讯录<二>(动态模拟)
- Spark Sort-based Shuffle
- Problem A(逆元) 2016"百度之星" - 资格赛(Astar Round1)
- iOS开发系列--Objective-C之KVC、KVO
- 【项目管理和构建】——Maven下载、安装和配置(二)
- linuxc编程一站式学习(c基础)
- Jquery 复习练习(02)Javascript 与jquery 互转 onclick 与click区别
- 【项目管理和构建】——Maven简介(一)
- yum 本地仓库配置(CentOS6.4系统)
- laravel学习之依赖注入
- 51nod 1040 最大公约数的和 欧拉函数
- NKOI 3679 找数2
- LeetCode-5.Longest Palindromic Substring
- poj 1163 The Triangle
- 二进制