A. Theatre Square time limit per test:1 second memory limit per test:256 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.

Examples Input 6 6 4 Output 4 題目鏈接：http://codeforces.com/problemset/problem/1/A 分析： 題意：給你一個矩形的常和寬，以及邊長為a的正方形磚塊，用磚塊去鋪這個矩形，允許重疊，不難，記住開__int64，否則會WA！ 下面給出AC代碼： 1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 __int64 n,m,a; 6 while(scanf("%I64d%I64d%I64d",&n,&m,&a)!=EOF) 7 { 8 __int64 t=n/a; 9 __int64 s=m/a; 10 if(n%a!=0) 11 t+=1; 12 if(m%a!=0) 13 s+=1; 14 __int64 k=t*s; 15 printf("%I64d\n",k); 16 } 17 return 0; 18 } B. Spreadsheets time limit per test：10 seconds memory limit per test：64 megabytes input：standard input output：standard output

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.

Input

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 .

Output

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

Examples Input 2
R23C55
BC23 Output BC23
R23C55 題目鏈接：http://codeforces.com/problemset/problem/1/B 分析：?????????? 題意：在Excel中，一個格子的位置有2種表示：

例如第23行第55列

①R23C55

②BC23

第一種表示方法很直觀。

第二種表示方法中BC表示列。23表示行。

1-26列：A, B, C...Z

27-?列：AA, AB, AC...AZ, BA, BB, BC...ZZ

?-?：AAA...ZZZ...

跟進制的轉換很類似！

輸入任意一種表示，你的任務是輸出另一種表示，模擬即可！
下面給出AC代碼： 1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn=1000005; 4 typedef long long ll; 5 char str[maxn]; 6 int flag; 7 void change(int n)//26進制轉換 8 { 9 if(n>26) 10 change((n-1)/26); 11 printf("%c",(n-1)%26+'A'); 12 } 13 void solve1() 14 { 15 int row=0; 16 int col=0; 17 for(int i=1;i<flag;i++) 18 if(isdigit(str[i])) 19 row=row*10+str[i]-'0';//計算行 20 for(int i=flag+1;str[i];i++) 21 col=col*10+str[i]-'0';//計算列 22 change(col); 23 printf("%d\n",row); 24 } 25 void solve2() 26 { 27 int row=0; 28 int col=0; 29 for(int i=0;str[i];i++) 30 { 31 if(isupper(str[i])) 32 col=col*26+str[i]-'A'+1;//計算列 33 else row=row*10+str[i]-'0';//計算行 34 } 35 printf("R%dC%d\n",row,col); 36 } 37 int main() 38 { 39 int T; 40 while(scanf("%d",&T)!=EOF) 41 { 42 while(T--) 43 { 44 scanf("%s",str); 45 flag=0; 46 if(str[0]=='R'&&isdigit(str[1])) 47 { 48 for(int i=2;str[i];i++) 49 { 50 if(str[i]=='C') 51 { 52 flag=i; 53 break; 54 } 55 } 56 } 57 if(flag) 58 solve1();//判斷‘R23C55’這一種情況 59 else 60 solve2();//判斷‘BC23’這一種情況 61 } 62 } 63 return 0; 64 } C. Ancient Berland Circus time limit per test：2 seconds memory limit per test：64 megabytes input：standard input output：standard output

Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples Input 0.000000 0.000000
1.000000 1.000000
0.000000 1.000000 Output 1.00000000 ??題目鏈接：http://codeforces.com/problemset/problem/1/C 分析： 題意：有一個正n邊形

輸入正n邊形的其中3個點

問正n邊形可能存在的最小面積，已知n<=100

該題關鍵技巧就是要畫外接圓，然后玩玩圓周角，圓心角這些概念，當個平面幾何問題，先盡量多推出一些結論。

具體解法如下:

首先，隨便畫個正多少邊形，畫個外接圓。根據正弦定理，可以直接知道外接圓半徑。把這三個點連成一個三角形，三個角都會是正x邊形的一個邊對應這個外接圓的圓周角的整數倍。由于x很小，枚舉+判斷就可以了。

三角形外接圓半徑公式：



每條邊所對應的圓心角 = 2*PI/n

所以圓周角 = 圓心角/2 = PI/n

正n邊形面積：

????????????? ?下面給出AC代碼：? 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 }

?

???????????

???????

???????????

??????? ???????

???????????

轉載于:https://www.cnblogs.com/ECJTUACM-873284962/p/6533637.html

總結

以上是生活随笔為你收集整理的Codeforces Beta Round #1 A,B,C的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。