Problem D: Division
时间： 1ms 内存：128M
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：
输出：
示例输入：
2 9 3
2 3 2
21 42 7
123 911 1
示例输出：
(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.
提示：
参考答案（内存最优[760]）：
#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("%0.2lf\n",area);
x = xx; y = yy;
}
}参考答案（时间最优[16]）：
#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("%0.2lf\n",area);
x = xx; y = yy;
}
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