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 ci…

http://codeforces.com/problemset/problem/1/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 si…

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'…

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 o…

A - Theatre Square Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 1A Description Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the…

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 ci…

题目链接:B.New Theatre Square 题意: 你要把所有"." 都变成"*",你可以有两个选择,第一种就是一次铺一个方块(1*1),第二种就是同一行一次铺两个(1*2).第一种花费x,第二种花费y.问最少花费多少能把所有铺完 题解: 如果y>=2*x,那么就直接找到所有".",然后乘于x就行 否则就找俩俩一对就行了 代码: 1 #include<stdio.h> 2 #include<algorithm>…

题目大意; 有一个广场,广场的大小是n*m,  有a*a的石砖,石砖铺广场可以比广场大,石砖不能切割.问最少需要多少个石砖. ===================================================================== #include <iostream> #include <cmath> #include <algorithm> #include <string> #include <deque>…

题目链接:http://www.codeforces.com/problemset/problem/1/A题意:至少用多少块边长为a的方块铺满NxM的矩形区域.C++代码: #include <iostream> using namespace std; int main() { long long n, m, a; cin >> n >> m >> a; cout << ((n+a-)/a) * ((m+a-)/a); ; } C++…

从今天開始.就要在Codeforces里有一个新的開始了,貌似任务非常重的说~~ Codeforces专题我将会记录全部通过的题目,事实上仅仅要通过的题目都是水题啊!. 题目大意: 依照要求计算须要多少flagstone. 以下是代码: #include <stdio.h> int main() { long long n,m,a,r,c; scanf("%I64d%I64d%I64d",&n,&m,&a); r=n/a; if(n%a)r++; c…