1113: Problem D: Division

题目描述

Problem D: Division
Given t, a, b positive integers not bigger than 2147483647, establish whether (t^a – 1)/(t^b -1) is an integer with less than 100 digits. Each line of input contains t, a, b. For each line of input print the formula followed by its value, or followed by "is not an integer with less than 100 digits.", whichever is appropriate.

输入

暂无

输出

暂无

样例输入

2 9 3
2 3 2
21 42 7
123 911 1

样例输出

(2^9-1)/(2^3-1) 73
(2^3-1)/(2^2-1) is not an integer with less than 100 digits.
(21^42-1)/(21^7-1) 18952884496956715554550978627384117011154680106
(123^911-1)/(123^1-1) is not an integer with less than 100 digits.

参考代码

#include <math.h>
linepoints(double x1, double y1, double x2, double y2,
           double *a, double *b, double *c) 
{
    *a = y2 - y1;
    *b = x1 - x2;
    *c = *a * x1 + *b * y1;
}
bi(double x1, double y1, double x2, double y2, double *a, double *b, double *c) 
{
    *a = 2*(x2-x1);
    *b = 2*(y2-y1);
    *c = x2*x2 + y2*y2 - x1*x1 - y1*y1;
}
isct(double a, double b, double c, double aa, double bb, double cc,
     double *x, double *y) 
{
    double det = a*bb - b*aa;
    if (fabs(det) < 1e-10) return 0;
    *x = (-b*cc + c*bb)/det;
    *y = (a*cc - c*aa)/det;
    return 1;
}
int nd, done[200];
int i,j,k;
struct pp 
{
    double a, b, c;
}
p[100];
double x,y,xx,yy,a,b,c;
int np, i,j,j;
char hc[20];
double xa, xb, ya, yb, area, triarea;
int nx;
main() 
{
    linepoints(0,0,0,10,&a,&b,&c);
    p[np].a = a;
    p[np].b = b;
    p[np++].c = c;
    linepoints(0,10,10,10,&a,&b,&c);
    p[np].a = a;
    p[np].b = b;
    p[np++].c = c;
    linepoints(10,10,10,0,&a,&b,&c);
    p[np].a = a;
    p[np].b = b;
    p[np++].c = c;
    linepoints(10,0,0,0,&a,&b,&c);
    p[np].a = a;
    p[np].b = b;
    p[np++].c = c;
    while (3 == scanf("%lf%lf %s",&xx,&yy,hc)) 
    {
        if (strcmp(hc,"Colder")) bi(x,y,xx,yy,&p[np].a,&p[np].b,&p[np].c),np++;
        if (strcmp(hc,"Hotter")) bi(xx,yy,x,y,&p[np].a,&p[np].b,&p[np].c),np++;
        area = 0;
        nd = 0;
        for (i=0;i<np;i++) 
        {
            nx = 0;
            for (j=0;j<np;j++) 
            {
                if (isct(p[i].a,p[i].b,p[i].c,p[j].a,p[j].b,p[j].c,&x,&y)) 
                {
                    int isin = 1;
                    for (k=0;k<np;k++) 
                    {
                        if (p[k].a*x + p[k].b*y - p[k].c < -.000001) isin = 0;
                    }
                    if (isin) 
                    {
                        if (!nx) 
                        {
                            nx++;
                            xa = x;
                            ya = y;
                        } else if (nx<2 && (fabs(x-xa)>.000001 || fabs(y-ya)>.000001)) 
                        {
                            nx++;
                            xb = x;
                            yb = y;
                        }
                    }
                }
            }
            if (nx == 2) 
            {
                for (j=0;j<nd;j++) 
                {
                    if (fabs(p[i].a*p[done[j]].b - p[i].b*p[done[j]].a) < .000001
                                     && fabs(p[i].a*p[done[j]].c - p[i].c*p[done[j]].a) < .000001
                                     && fabs(p[i].b*p[done[j]].c - p[i].c*p[done[j]].b) < .000001)
                                     break;
                }
                if (j == nd) 
                {
                    done[nd++] = i;
                    triarea = -p[i].c / hypot(p[i].a,p[i].b) * hypot(xa-xb,ya-yb) / 2;
                    area += triarea;
                }
            }
        }
        if (nd < 3 || area < .000001) area = 0;
        printf("%0.2lfn",area);
      x = xx; y = yy;
   }
}

解析

暂无