(解题报告)Theatre Square
2015-11-30 19:04
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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
一道简单的求最小瓷砖数的题!分为长和宽依次除以瓷砖的边长,能整除就是其对应的瓷砖数,否则就是瓷砖数加1;最后将长和宽各自的瓷砖数加起来就得到了总的瓷砖数;
代码如下:
仅代表个人观点,不喜勿喷,欢迎交流!
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
一道简单的求最小瓷砖数的题!分为长和宽依次除以瓷砖的边长,能整除就是其对应的瓷砖数,否则就是瓷砖数加1;最后将长和宽各自的瓷砖数加起来就得到了总的瓷砖数;
代码如下:
#include <stdio.h> int main() { __int64 m,n,a,l1,l2; scanf("%I64d%I64d%I64d",&m,&n,&a); // 注意是大写的%I64d if(m % a == 0) { l1 = m / a; } else { l1 = m / a + 1; } if(n % a == 0) { l2 = n / a; } else { l2 = n / a + 1; } printf("%I64d",l1 * l2); return 0; }
仅代表个人观点,不喜勿喷,欢迎交流!
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