Codeforces Beta Round #1

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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.

Sample test(s)
input
6 6 4
output
4

题意：给你一个矩形的常和宽，以及边长为a的正方形砖块，用砖块去铺这个矩形，允许重叠，直接用ceil函数

#include <cstdio>#include <cmath>int main(){    int m,n,a;    scanf("%d%d%d",&m,&n,&a);    printf("%.lf\n",ceil(m*1.0/a)*ceil(n*1.0/a));    return 0;}

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.

Sample test(s)
input
2R23C55BC23
output
BC23R23C55

题意：类似于Excel里面的表示方法，用来转换两种表示方法之间的编码，直接暴力。。。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <stack>using namespace std;char s[20];bool judge(char s[]){    int l=strlen(s);    int ans=0;    for(int i=1;i<l;i++)        if((s[i]>='0'&&s[i]<='9')&&(s[i-1]>='A'&&s[i-1]<='Z'))            ans++;    if(ans==2)        return true;    return false;}int main(){    int test;    //freopen("a.txt","r",stdin);    //freopen("b.txt","w",stdout);    scanf("%d",&test);    while(test--)    {        scanf("%s",s);        if(judge(s))        {            int l=strlen(s);            int r=0,c=0;            int rr=1,cc=1;            int i,j;            for(i=l-1;i>=0;i--)            {                if(s[i]>='A'&&s[i]<='Z')                    break;                if(s[i]>='0'&&s[i]<='9')                {                    c+=cc*(s[i]-'0');                    cc*=10;                }            }/*            for(;i>=0;i--)            {                if(s[i]>='A'&&s[i]<='Z')                    continue;                if(s[i]>='0'&&s[i]<='9')                    break;            }            i++;*/            for(j=i-1;j>=0;j--)            {                if(s[j]>='A'&&s[j]<='Z')                    break;                if(s[j]>='0'&&s[j]<='9')                {                    r+=rr*(s[j]-'0');                    rr*=10;                }            }            stack<char>st;            while(!st.empty())                st.pop();            while(c)            {                int num=c%26;                if(num==0)                {                    st.push('Z');                    c--;                }                else                    st.push(num-1+'A');                c/=26;            }            while(!st.empty())            {                printf("%c",st.top());                st.pop();            }            printf("%d\n",r);        }        else        {            int r=0,c=0;            int rr=1,cc=1;            int i,j,l=strlen(s);            stack<char>st;            stack<char>nt;            while(!nt.empty())                nt.pop();            while(!st.empty())                st.pop();            for(i=0;i<l;i++)            {                if(s[i]>='0'&&s[i]<='9')                    break;                st.push(s[i]);            }            for(j=i;j<l;j++)            {                nt.push(s[j]);            }            while(!st.empty())            {                c+=cc*(st.top()-'A'+1);                cc*=26;                st.pop();            }            while(!nt.empty())            {                r+=rr*(nt.top()-'0');                rr*=10;                nt.pop();            }            printf("R%dC%d\n",r,c);        }    }    return 0;}

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.

Sample test(s)
input
0.000000 0.0000001.000000 1.0000000.000000 1.000000
output
1.00000000

题意：给你三个点的坐标，求由这三个点得到的正多边形的面积，由于不知道这个正多边形有多少条边，以及不清楚由这n个等腰三角形的每个三角形的面积，开始我想的方法是求由这三个点组成的三角形的外接圆及半径，然后每条边于外心组成的三角形中圆心角最小的那个三角形对应的那条边应该就是这个正多边形组成的基本元素，然后WA了，原因是（思路错了!）

换种方法做，假使开始是3边形，这种与圆心组成的角中，如果每个角都满足能够满足这些角围成一圈就退出，直到n边形为止，这样一直枚举下去，找到合适的情况就break

// WA 了，思路跪了（感谢woshiren01指出错误）所以跪了#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define PI 3.141592653589793struct point{    double x;    double y;};point ans,p[3];struct Line{    point a;    point b;    double distance;    double jiao;};point intersection(Line u,Line v){    point res=u.a;    double k=((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));    res.x+=(u.b.x-u.a.x)*k;    res.y+=(u.b.y-u.a.y)*k;    return res;}point circumcenter(point a,point b,point c){    Line u,v;    u.a.x=(a.x+b.x)/2;    u.a.y=(a.y+b.y)/2;    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)/2;    v.a.y=(a.y+c.y)/2;    v.b.x=v.a.x-a.y+c.y;    v.b.y=v.a.y+a.x-c.x;    return intersection(u,v);}bool cmp(Line xx,Line yy){    return xx.jiao>yy.jiao;}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 cal(Line ll){    double xx=dis(ll.a,ans);    double yy=dis(ll.b,ans);    double zz=ll.distance;    return (xx*xx+yy*yy-zz*zz)/2/xx/yy;}double area(double x0,double y0,double x1,double y1,double x2,double y2){    return (x0*y1+x2*y0+x1*y2-x2*y1-x0*y2-x1*y0)/2.0;}int main(){    for(int i=0;i<3;i++)        scanf("%lf%lf",&p[i].x,&p[i].y);    ans=circumcenter(p[0],p[1],p[2]);    Line x[3];    x[0].distance=dis(p[1],p[2]);    x[1].distance=dis(p[0],p[2]);    x[2].distance=dis(p[0],p[1]);    x[0].a=p[1],x[0].b=p[2];    x[1].a=p[0],x[1].b=p[2];    x[2].a=p[0],x[2].b=p[1];    for(int i=0;i<3;i++)        x[i].jiao=cal(x[i]);    sort(x,x+3,cmp);    //printf("%lf....%lf\n",x[0].a.x,x[0].a.y);    //printf("%lf....%lf\n",x[0].b.x,x[0].b.y);    //printf("%lf\n",acos(x[0].jiao));    double ccc=2*PI/acos(x[0].jiao);    //printf("%lf\n",ccc);    double mmmm=ccc*area(ans.x,ans.y,x[0].a.x,x[0].a.y,x[0].b.x,x[0].b.y);    mmmm<0?(mmmm*=-1):1;    printf("%.8lf\n",mmmm);    return 0;}

换方法之后，果然过了。

#include <cstdio>#include <iostream>#include <cmath>#include <algorithm>using namespace std;#define pi 3.1415926#define eps 10e-3typedef struct point{    double x,y;};double area(double x0,double y0,double x1,double y1,double x2,double y2){    return 0.5*fabs(x0*y1+x2*y0+x1*y2-x2*y1-x0*y2-x1*y0);}double dis(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}point p[3];bool is_ok(int n,double jiao){    double cnt=n*jiao/pi;    double pnt=floor(cnt+eps);    if(cnt-pnt<eps)        return true;    return false;}int main(){    for(int i=0;i<3;i++)        scanf("%lf%lf",&p[i].x,&p[i].y);    double a=dis(p[0],p[1]);    double b=dis(p[0],p[2]);    double c=dis(p[1],p[2]);    double A=acos((b*b+c*c-a*a)/2/b/c);    double B=acos((a*a+c*c-b*b)/2/a/c);    double C=acos((b*b+a*a-c*c)/2/b/a);    double r=a*b*c/area(p[0].x,p[0].y,p[1].x,p[1].y,p[2].x,p[2].y)/4;    //double jiao=max(A,max(B,C)); 开始我觉得下面的is_ok函数只要判断jiao就可以了// 其实还是会WA 第五组数据。。。WA的时候答案是输出的两倍，Wa的厉害了    int n;    for(n=3;n<=100;n++)        if(is_ok(n,A)&&is_ok(n,B)&&is_ok(n,C))            break;    double tag=2*pi/n;    double Area=n*r*r*sin(tag)/2;    printf("%.8lf\n",Area);    return 0;}