[Codeforces 系列] Codeforces 1A.Theatre Square

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解决一个数学问题，关于如何用特定大小的正方形砖块，最少数量地铺设一个矩形的剧院广场，确保没有砖块被切割且完全覆盖整个区域。

A. Theatre Square

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 ( $1 ≤ n, m, a ≤ 10^9$ ).

Output

Write the needed number of flagstones.

Examples

input

6 6 4

output

题意

有一个 $n * m$ 的矩形，要用 $a * a$ 的正方形来进行覆盖，正方形不能分割，求需要多少个正方形?

分析

问题很明确，由于正方形不能分割，则正方形覆盖面积一定大于等于矩形，我们将正方形从左到右从上到下放置，显然横着排列需要 $(n+a-1)/a$ (c 语言会自动取整，这里利用了一个小技巧)。同理得一共需要 $(m+a-1)/a$ 行，所以最终答案即为 $(long long)((n+a-1)/a) * (long long)((m+a-1)/a)$ 注意运算顺序和类型转换，同时注意到该数字范围会超过int，所以记得使用longlong类型。

代码

#include <cstdio>
int main()
{
    int n, m, a;
    scanf("%d%d%d", &n, &m, &a);
    printf("%I64d", (long long)((n - 1 + a) / a) * (long long)((m - 1 + a) / a));
    return 0;
}