Codeforces Beta Round #1

A题，水题。

B题也是水题，弄的比较麻烦，前几天队内赛见过，水题怎么水都能过。

C题

题意：给出正n边形上的三个点，求最小的正n边形的面积。

以前貌似见过此题。思路也没什么进展，就是枚举n，通过旋转a，判断b c是否在多边形上。感觉是水过的，改了改eps就过了，看别人代码，还有神奇的gcd的做法。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int s[];
int que[];
void judge(int x)
{
    int num,n,i;
    for(i = ;i <= ;i ++)
    {
        if(s[i] >= x)
        break;
        x -= s[i];
    }
    memset(que,,sizeof(que));
    n = i;
    num = ;
    x--;
    while(x)
    {
        que[num++] = x%;
        x /= ;
    }
    for(i = n-;i >= ;i --)
    {
        printf("%c",que[i]+'A');
    }
    return ;
}
void fun1(char *str)
{
    int a,b,i;
    a = b = ;
    int len;
    len = strlen(str);
    for(i = ;i < len;i ++)
    {
        if(str[i] == 'C') break;
        a = a* + (str[i]-'');
    }
    for(i = i+;i < len;i ++)
    b = b* + (str[i]-'');
    judge(b);
    printf("%d\n",a);
}
void fun2(char *str)
{
    int a = ,i,j;
    int len;
    len = strlen(str);
    for(i = ;i < len;i ++)
    {
        if(str[i] <= ''&&str[i] >= '')
        break;
    }
    int num = ;
    for(j = i-;j >= ;j --)
    a += s[num++]*(str[j] - 'A' + );
    printf("R");
    for(;i < len;i ++)
    printf("%c",str[i]);
    printf("C%d\n",a);
}
int main()
{
    int t,len,i;
    char str[];
    scanf("%d",&t);
    s[] = ;
    for(i = ;i <= ;i ++)
    s[i] = s[i-]*;
    while(t--)
    {
        scanf("%s",str);
        len = strlen(str);
        if(str[] == 'R')
        {
            int flag = ,z = ;
            for(i = ;i < len;i ++)
            {
                if(str[i] == 'C'&&i >= )
                flag ++;
                else if(str[i] <= 'Z'&&str[i] >= 'A')
                z = ;
            }
            if(flag == &&z == )
            fun1(str);
            else
            fun2(str);
        }
        else
        fun2(str);
    }
    return ;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
using namespace std;
#define PI 3.1415926
#define eps 1e-3
struct point
{
    double x,y;
};
struct line
{
    point a,b;
};

point intersection(line u,line v)
{
    point ret = u.a;
    double t = ((u.a.x-v.a.x)*(v.a.y-v.b.y) - (u.a.y-v.a.y)*(v.a.x-v.b.x))
    /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x += (u.b.x-u.a.x)*t;
    ret.y += (u.b.y-u.a.y)*t;
    return ret;
}
point circumcenter(point a,point b,point c)
{
    line u,v;
    u.a.x = (a.x+b.x)/;
    u.a.y = (a.y+b.y)/;
    u.b.x = u.a.x-a.y+b.y;
    u.b.y = u.a.y+a.x-b.x;
    v.a.x = (a.x + c.x)/;
    v.a.y = (a.y + c.y)/;
    v.b.x = v.a.x - a.y + c.y;
    v.b.y = v.a.y + a.x - c.x;
    return intersection(u,v);
}
point Rotate(point p,double angle)
{
    point res;
    res.x = p.x * cos(angle) - p.y*sin(angle);
    res.y = p.x * sin(angle) + p.y*cos(angle);
    return res;
}
int judge(point a,point b)
{
    double t;
    t = a.x - b.x;
    if(t < )
    t = -t;
    if(t > eps) return ;
    t = a.y - b.y;
    if(t < )
    t = -t;
    if(t > eps) return ;
    return ;
}
int main()
{
    int i,j,n;
    point a,b,c,r,d;
    point p[];
    scanf("%lf%lf",&a.x,&a.y);
    scanf("%lf%lf",&b.x,&b.y);
    scanf("%lf%lf",&c.x,&c.y);
    r = circumcenter(a,b,c);
    d.x = a.x - r.x;
    d.y = a.y - r.y;
    for(i = ;i <= ;i ++)
    {
        int num = ;
        p[] = a;
        for(j = ;j < i;j ++)
        {
            p[j] = Rotate(d,*PI*j/i);
            p[j].x += r.x;
            p[j].y += r.y;
            if(judge(p[j],b))
            num ++;
            if(judge(p[j],c))
            num ++;
        }
        if(num == ) break;
    }
    n = i;
    double s1,s2;
    s1 = s2 = ;
    for(i = ;i < n;i ++)
    {
        s1 += p[(i+)%n].y*p[i].x;
        s2 += p[(i+)%n].y*p[(i+)%n].x;
    }
    printf("%.7lf",fabs(s1-s2)/);
    return ;
}