6 6 4

4

In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.

The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.

Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.

Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.

The first line of the input contains integer number n (1 ≤ n ≤ 10⁵), the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 10⁶ .

Write n lines, each line should contain a cell coordinates in the other numeration system.

0.000000 0.000000 1.000000 1.000000 0.000000 1.000000

1.00000000

题目链接：http://codeforces.com/problemset/problem/1/C

输入正n边形的其中3个点

问正n边形可能存在的最小面积，已知n<=100

该题关键技巧就是要画外接圆，然后玩玩圆周角，圆心角这些概念，当个平面几何问题，先尽量多推出一些结论。

具体解法如下:

首先，随便画个正多少边形，画个外接圆。根据正弦定理，可以直接知道外接圆半径。把这三个点连成一个三角形，三个角都会是正x边形的一个边对应这个外接圆的圆周角的整数倍。由于x很小，枚举+判断就可以了。

三角形外接圆半径公式：

每条边所对应的圆心角 = 2*PI/n

所以圆周角 = 圆心角/2 = PI/n

正n边形面积：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const double PI=3.1415926535;
 4 const double ERR=0.01;
 5 bool feq(double a,double b)
 6 {
 7     return fabs(a-b)<ERR;
 8 }
 9 double fgcd(double a,double b)
10 {
11     if (feq(a,0))
12         return b;
13     if (feq(b,0))
14         return a;
15     return fgcd(b,fmod(a,b));
16 }
17 double dist(double x0,double x1,double y0,double y1)
18 {
19     return sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0));
20 }
21 int main()
22 {
23     double x[3];
24     double y[3];
25     for (int i=0;i<3;i++)
26         scanf("%lf%lf",&x[i],&y[i]);
27     double l[3];
28     for (int i=0;i<3;i++) l[i]=dist(x[i],x[(i+1)%3],y[i],y[(i+1)%3]);
29     double p=(l[0]+l[1]+l[2])/2;
30     double s=sqrt(p*(p-l[0])*(p-l[1])*(p-l[2]));
31     double r=l[0]*l[1]*l[2]/(s*4);
32     double ang[3];
33     for (int i=0;i<3;i++) ang[i] = acos(1-l[i]*l[i]/(2*r*r));
34     ang[2] =2*PI-ang[0]-ang[1];
35     double unita =0;
36     for (int i=0;i<3;i++)
37         unita=fgcd(unita,ang[i]);
38     printf("%.6lf\n",r*r*sin(unita)*PI/unita);
39     return 0;
40 }