Codeforces Beta Round #1

总的来说因为是第一期所以感觉上还是难度很低 cf的英文题确实比国内一些OJ层次高一点第一题水水过去但是二三题因为题意耽误了很多时间第二题因为弄错意思WA了两次第三题感觉数学还需要加强总体来说第一次感觉还不错抽出时间复习的情况下还是完成了所有题目并且整理了博客下周考试前再做两次模拟希望状态会好一点

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

#include <cstdio> #include <cmath> typedef long long LL; int main(){ LL n, m, a; scanf("%I64d%I64d%I64d", &n, &m, &a); printf("%I64d\n", LL(ceil(1.0*n/a) * ceil(1.0*m/a))); return 0; }

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <stack> #include <queue> using namespace std; char str[1010]; char ch[22]; int main(){ int n; scanf("%d", &n); while(n--){ scanf("%s", str); bool flag = false; for(int i = 2; i < strlen(str); ++i){ if(str[0] == 'R' && str[1] - '0' < 10 && (str[i] == 'C')){ //判断是第一种形式还是第二种形式 flag = true; break; } } if(flag){ bool fi = true; //标记行和列的读取 int a = 0, b = 0; for(int i = 1; i < strlen(str); ++i){ if(str[i] == 'C'){ //开始去读列 fi = false; continue; } if(fi){ //行转换数字 a = a * 10 + (str[i] - '0'); } else{ //列转换数字 b = b * 10 + (str[i] - '0'); } } int k = 0; while(b){ int c = b % 26; //转换成相应字母 b /= 26; if(c == 0){ c += 26; b -= 1; } ch[k++] = c + 'A' - 1; } for(int i = k-1; i >= 0; --i){ printf("%c", ch[i]); } printf("%d\n", a); } else{ bool fi = true; int a = 0, b = 0; for(int i = 0; i < strlen(str); ++i){ if(str[i] - '0' < 10){ //开始去读列 fi = false; } if(fi){ a = a * 26 + (str[i] + 1 - 'A'); //转换数字 } else{ b = b * 10 + (str[i] - '0'); //转换数字 } } printf("R%dC%d\n", b, a); } } return 0; }

#include <iostream> #include <cstdio> #include <cmath> using namespace std; const double PI = 3.1415926535; const double ERR = 0.01; double dist(double x0, double x1, double y0, double y1); bool feq(double a, double b); double fgcd(double a, double b); int main(){ pair<double, double> po[3]; for(int i = 0; i < 3; ++i){ scanf("%lf%lf", &po[i].first, &po[i].second); } double len[3]; for(int i = 0; i < 3; ++i){ len[i] = dist(po[i].first, po[(i+1)%3].first, po[i].second, po[(i+1)%3].second); } double p = (len[0] + len[1] + len[2]) / 2; //海伦公式: p=(a+b+c)/2 double s = sqrt(p * (p-len[0]) * (p-len[1]) * (p-len[2]));//海伦公式: S=√p(p-a)(p-b)(p-c) double r = len[0] * len[1] * len[2] / (s * 4); //求外接圆半径r=abc/4S double ang[3]; for(int i = 0; i < 3; ++i){ //由余弦定理求出三个圆心角ang[3] ang[i] = acos(1 - len[i]*len[i] / (2 * r * r)); } ang[2] = 2 * PI - ang[0] - ang[1]; //有可能有三个点在同一段半圆弧上,这是第三个圆心角应该用2π-ang[0]-ang[1] double unita = 0; for(int i = 0; i < 3; ++i){ //求这三个角的最大公约数为unita, 那么这就是一个正2π/unita边形 unita = fgcd(unita, ang[i]); } printf("%.6lf\n", r * r * sin(unita) * PI / unita); //S=1/2·r * r * sinA * n return 0; } double dist(double x0, double x1, double y0, double y1){ return sqrt((x1-x0) * (x1-x0) + (y1-y0) * (y1-y0)); } bool feq(double a, double b){ return fabs(a-b) < ERR; } double fgcd(double a, double b){ if(feq(a, 0)){ return b; } if(feq(b, 0)){ return a; } return fgcd(b, fmod(a, b)); }