Codeforces Beta Round #1

A:大水题；

代码：

#include<iostream>
#define ll long long
using namespace std;

int main()
{
    ll n,m,a;
    cin>>n>>m>>a;
    cout<<(((n+a-1)/a)*((m+a-1)/a));
    return 0;
}

B：灵活的运用sscanf

代码：

#include<cstdio>
#include<cstring>
#define maxn 1000005
using namespace std;
int n;
char s[50];
void a_to_b()
{
    int a,b;
    int p=maxn-1;
    char t[maxn];
    sscanf(s,"R%dC%d",&a,&b);
    t[p--]=0;
    while(a)
    {
        t[p--]=a%10+'0';
        a/=10;
    }
    while(b)
    {
        t[p--]=(b-1)%26+'A';
        b=(b-1)/26;
    }
    printf("%s
",&t[p+1]);
}

void b_to_a()
{
    char t[maxn];
    int c = 0;
    int r;
    sscanf(s,"%[A-Z]%d",t,&r);
    int len = strlen(t);
    int p = 0;
    while (p<len)
    {
        c+=t[p++]-'A'+1;
        c*=26;
    }
    c/=26;
    printf("R%dC%d
",r,c);
}

int main()
{
    int a,b;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",s);
        if(sscanf(s,"R%dC%d",&a,&b)==2)a_to_b();
        else b_to_a();
    }
    return 0;
}

C：

很水的一个计算几何，关键是枚举，因为最多100个角；

稍微要用到一点圆的知识，因为它是正的多边形，所以所有的点都在外接圆上，判断三角形的内角是否是单位圆心角一半的倍数；

代码：

#include<cstdio>
#include<cmath>
#define eps 0.0001
#define pi acos(-1)
using namespace std;

struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y){ }
}p[3];
point operator - (point a,point b){return point(a.x-b.x,a.y-b.y);}
double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}

bool isok(double a,int n)
{
    double alph=a*n/pi;
    double x=floor(alph+eps);
    if(alph-x<eps)return 1;
    return 0;
}

int main()
{
    for(int i=0;i<3;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
    double d1=dis(p[0],p[2]);
    double d2=dis(p[1],p[0]);
    double d3=dis(p[2],p[1]);
    double a=asin(cross(p[1]-p[0],p[2]-p[0])/d1/d2);
    double b=asin(cross(p[2]-p[1],p[0]-p[1])/d2/d3);
    double c=asin(cross(p[0]-p[2],p[1]-p[2])/d3/d1);
    double r=d1*d2*d3/2/cross(p[1]-p[0],p[2]-p[0]);
    int i;
    for(i=3;i<=100;i++)
        if(isok(a,i)&&isok(b,i)&&isok(c,i))break;
    double ans=i*r*r*sin(2*pi/i)*0.5;
    printf("%lf",ans);
}