几何公式

两条直线是否相交

 1 //叉积
 2 double mult(Point a, Point b, Point c)
 3 {
 4     return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
 5 }
 6
 7 //aa, bb为一条线段两端点 cc, dd为另一条线段的两端点 相交返回true, 不相交返回false
 8 bool intersect(Point aa, Point bb, Point cc, Point dd)
 9 {
10     if ( max(aa.x, bb.x)<min(cc.x, dd.x) )
11     {
12         return false;
13     }
14     if ( max(aa.y, bb.y)<min(cc.y, dd.y) )
15     {
16         return false;
17     }
18     if ( max(cc.x, dd.x)<min(aa.x, bb.x) )
19     {
20         return false;
21     }
22     if ( max(cc.y, dd.y)<min(aa.y, bb.y) )
23     {
24         return false;
25     }
26     if ( mult(cc, bb, aa)*mult(bb, dd, aa)<0 )
27     {
28         return false;
29     }
30     if ( mult(aa, dd, cc)*mult(dd, bb, cc)<0 )
31     {
32         return false;
33     }
34     return true;
35 }

圆周率计算代码

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int n;
 5     while(scanf("%d",&n)!=EOF&&n!=0)
 6     {
 7         int const N=7200;
 8         int const M=10000;
 9         int const  B=10000;
10         int const L=4;
11         int s[M/L];
12         int r1[N]= {0},r2[N]= {0},d1[N]= {0},d2;
13         int r3[N]= {0},r4[N]= {0},d3[N]= {0},d4;
14         int i,k,t,p=0,mp=M/L/20;
15         r1[0]=1;
16         r1[1]=3;
17         r3[0]=4;
18         printf("正在计算,请等待\n____________________\n");
19         for(k=0; k<M/L; ++k)
20         {
21             t=r1[0]*B;
22             d1[0]=t/0x5;
23             r1[0]=t%0x5;
24 //
25             t=r3[0]*B;
26             d3[0]=t/0xEF;
27             r3[0]=t%0xEF;
28             s[k]=d1[0]-d3[0];
29             int tag=0;
30             for(i=1; i<N; ++i)
31             {
32                 t=r1[i]*B+d1[i-1];
33                 d1[i]=t/0x19;
34                 r1[i]=t%0x19;
35                 t=r2[i]*B+d1[i];
36                 d2=t/(2*i+1);
37                 r2[i]=t%(2*i+1);
38 //
39                 t=r3[i]*B+d3[i-1];
40                 d3[i]=t/0xDF21;
41                 r3[i]=t%0xDF21;
42                 t=r4[i]*B+d3[i];
43                 d4=t/(2*i+1);
44                 r4[i]=t%(2*i+1);
45                 if(tag)
46                 {
47                     s[k]+=(d2-d4);
48                     tag=0;
49                 }
50                 else
51                 {
52                     s[k]+=(d4-d2);
53                     tag=1;
54                 }
55             }
56             if(p==mp)
57             {
58                 printf(">");
59                 p=0;
60             }
61             else
62                 p++;
63         }
64         for(i=M/L-1; i>=0; i--)
65         {
66             while(s[i]>=B)
67             {
68                 s[i-1]++;
69                 s[i]-=B;
70             }
71             while(s[i]<0)
72             {
73                 s[i-1]--;
74                 s[i]+=B;
75             }
76         }
77         printf("\npi=3.\n");
78         for(i=0; i<M/L; ++i)
79             printf("%d",s[i]);
80         printf("\n");
81     }
82     return 0;
83
84 }

求圆心

 1 void count(double x1,double y1,double x2,double y2,double x3,double y3)
 2 {
 3     double a,b,c,d,e,f;
 4     a=2*(x2-x1);
 5     b=2*(y2-y1);
 6     c=x2*x2+y2*y2-x1*x1-y1*y1;
 7     d=2*(x3-x2);
 8     e=2*(y3-y2);
 9     f=x3*x3+y3*y3-x2*x2-y2*y2;
10     x=(b*f-e*c)/(b*d-e*a);
11     y=(d*c-a*f)/(b*d-e*a);
12     r=sqrt((x-x1)*(x-x1)+(y-y1)*(y-y1));
13
14 }