Geometric template of algorithm Summary

1. geometric formula

Triangle:

1. Half perimeter P = (A + B + C)/2

2. area S = AHA/2 = absin (C)/2 = SQRT (P-A) (p-B) (p-C ))

3. midline MA = SQRT (2 (B ^ 2 + C ^ 2)-a ^ 2)/2 = SQRT (B ^ 2 + C ^ 2 + 2 bccos ()) /2

4. ta = SQRT (BC (B + C) ^ 2-A ^ 2)/(B + C) = 2 bccos (A/2)/(B + C)

5. california HA = bsin (c) = Csin (B) = SQRT (B ^ 2-(a ^ 2 + B ^ 2-C ^ 2)/(2a) ^ 2)

6. incircle radius R = s/P = Asin (B/2) sin (C/2)/sin (B + C)/2)

= 4 rsin (A/2) sin (B/2) sin (C/2) = SQRT (P-A) (p-B) (p-C) /P)

= Ptan (A/2) Tan (B/2) Tan (C/2)

7. outer circle radius R = ABC/(4S) = A/(2sin (A) = B/(2sin (B) = C/(2sin (c ))

Quadrilateral:

D1 and D2 are diagonal lines, M is a diagonal point line, and a is a diagonal angle.

1. A ^ 2 + B ^ 2 + C ^ 2 + d ^ 2 = D1 ^ 2 + D2 ^ 2 + 4 m ^ 2

2. S = d1d2sin (a)/2

(The inner Quadrilateral of the following circle)

3. AC + bd = d1d2

4. S = SQRT (P-A) (p-B) (p-C) (p-D), P is the semi-perimeter

Positive n edge shape:

R indicates the outer circle radius, and r indicates the incircle radius.

1. center angle A = 2PI/n

2. Inner angle c = (n-2) PI/n

3. side length a = 2 SQRT (R ^ 2-r ^ 2) = 2 rsin (A/2) = 2 rtan (A/2)

4. area S = NAR/2 = nR ^ 2tan (A/2) = nR ^ 2sin (a)/2 = Na ^ 2/(4tan (A/2 ))

Circle:

1. arc length L = Ra

2. String Length A = 2 SQRT (2hr-h ^ 2) = 2 rsin (A/2)

3. Bow Height H = r-SQRT (R ^ 2-A ^ 2/4) = r (1-cos (A/2) = atan (A/4)/2

4. Slice area S1 = RL/2 = R ^ 2a/2

5. Bow area S2 = (RL-A (R-h)/2 = R ^ 2 (a-sin (A)/2

Prism:

1. Volume V = ah, A is the bottom area, H is high

2. The side area S = LP, L is the rib length, and P is the straight section perimeter.

3. Full area T = S + 2a

Pyramid:

1. Volume V = Ah/3, a is the bottom area, H is high

(The following is a positive pyramid)

2. side area S = LP/2, L is oblique height, p is the bottom perimeter

3. Full area T = S +

Pyramid:

1. Volume V = (A1 + A2 + SQRT (a1a2) h/3, a1.a2 is the upper and lower bottom area, H is high

(The following is the right pyramid)

2. side area S = (P1 + p2) l/2. p1.p2 indicates the perimeter of the upper and lower sides, and l indicates the oblique height.

3. Full area T = S + A1 + A2

Cylindrical:

1. side area S = 2 pirh

2. Full area T = 2pir (H + r)

3. Volume V = PIR ^ 2 h

Cone:

1. Bus L = SQRT (H ^ 2 + R ^ 2)

2. side area S = pirl

3. Full area T = PIR (L + r)

4. Volume V = PIR ^ 2 H/3

Table:

1. Bus L = SQRT (H ^ 2 + (r1-r2) ^ 2)

2. side area S = Pi (R1 + R2) L

3. Full area T = pir1 (L + R1) + pir2 (L + R2)

4. Volume V = Pi (R1 ^ 2 + R2 ^ 2 + r1r2) h/3

Ball:

1. Full area T = 4pir ^ 2

2. Volume V = 4pir ^ 3/3

Arena:

1. side area S = 2 pirh

2. Full area T = Pi (2RH + R1 ^ 2 + R2 ^ 2)

3. Volume V = PIH (3 (R1 ^ 2 + R2 ^ 2) + H ^ 2)/6

Ball slice:

1. Full area T = PIR (2 H + R0), h indicates the height of the Ballon crown, and R0 indicates the radius of the bottom of the Ballon crown.

2. Volume V = 2pir ^ 2 H/3

Binary geometric template

1. Point struct

1 struct point2 {3     double x;4     double y;5 }

2. Crossover:

1/* P0 is a common vertex */2 double xmult (point P1, point P2, point P0) {3 return (p1.x-forwarded X) * (p2.y-forwarded y) -(p2.x-Snapshot X) * (p1.y-Snapshot y); 4}

3. Two-Point distance:

1 double dis(point p1,point p2){2     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));3 }

4. Graham algorithm for the sorting of convex packet points CMP:

1 bool cmp(point a,point b)2 {3     if (CrossMul(a,b,p[1])>0 ||4             (CrossMul(a,b,p[1])==0&&dis(a,p[1])<dis(b,p[1]))) return true;5     return false;6 }

5. convex hull: 1) distance between two points 2) area 3) perimeter

1 int main () 2 {3 int N, I, j, top; 4 While (CIN> N) 5 {6 for (I = 1; I <= N; I ++) 7 {8 CIN> P [I]. x> P [I]. y; 9 If (P [I]. Y <p [1]. Y | (P [I]. y = P [1]. Y & P [I]. x <p [1]. x) 10 swap (P [1], p [I]); 11} 12 sort (p + 2, P + n + 1, CMP); 13 Top = 0; 14 Q [++ top] = P [1]; 15 Q [++ top] = P [2]; 16 for (I = 3; I <= N; I ++) 17 {18 while (top> 1 & crossmul (Q [Top], p [I], Q [Top-1]) <0) Top --; 19 Q [++ top] = P [I]; 20} 21/* convex hull diameter */22 int max = int_min; 23 for (INT I = 1; I <= top; I ++) 24 for (Int J = 1; j <I; j ++) 25 max = max> dis_2 (Q [I], Q [J])? MAX: dis_2 (Q [I], Q [J]); 26 printf ("% d \ n", max ); 27/* Convex Hull area */28 int area = 0; 29 for (INT I = 2; I <= top; I ++) 30 area + = crossmul (Q [I], Q [I + 1], Q [1]); 31 printf ("% d \ n", ABS (area )); 32/* convex hull perimeter */33 double sum = DIS (Q [1], Q [2]); 34 for (I = 3; I <= top; I ++) 35 sum + = DIS (Q [I], Q [I-1]); 36 sum + = DIS (Q [1], Q [Top]); 37 printf ("% d", sum); 38} 39 return 0; 40}

 

(To be continued)