Codeforces 1A-Theatre Square(水)

A. Theatre Square
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
周末的手速赛?。。没做 周末一直在做网络赛。。回来补一下
题意:给n*m的矩形,边长为a的正方形,求最少需要多少块正方形可以铺满矩形。。太水了,不说了
#include <iostream>
#include <vector>
#include <stack>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int INF=0x3f3f3f3f;
#define LL long long
int main()
{
	int n,m,a;
	while(cin>>n>>m>>a){
		LL x=n%a==0?n/a:n/a+1;
		LL y=m%a==0?m/a:m/a+1;
		cout<<x*y<<endl;
	}
	return 0;
}

### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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