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 ;
}