下圖有多少個三角形？

01-16

下圖有多少個三角形

數了半天,受不鳥了,還是用程序吧.

參考微博里的27個三角形是怎麼數出來的？ - 葉飛影的回答中的程序代碼.

struct Segment { char start; // 線段起點 char end; // 線段終點 char point1; // 線段經過的點 char point2; // 線段經過的點 };


void main()

{

    const int segments_count = 35;
    Segment segments[segments_count] =

    {

        {"A", "B", 0, 0},

        {"A", "C", "G", "H"},

        {"A", "D", "F", "J"},

        {"A", "E", 0, 0},

        {"A", "F", 0, 0},

        {"A", "G", 0, 0},

        {"A", "H", "G", 0},

        {"A", "J", "F", 0},     //8
        {"B", "C", 0, 0},

        {"B", "D", "H", "J"},

        {"B", "E", "F", "G"},

        {"B", "F", "G", 0},

        {"B", "G", 0, 0},

        {"B", "H", 0, 0},

        {"B", "I", "H", 0},     // 7
        {"C", "D", 0, 0},

        {"C", "E", "I", "J"},

        {"C", "G", "H", 0},

        {"C", "H", 0, 0},

        {"C", "I", 0, 0},

        {"C", "J", "I", 0},     //6
        {"D", "E", 0, 0},

        {"D", "F", "J", 0},

        {"D", "H", "I", 0},

        {"D", "I", 0, 0},

        {"D", "J", 0, 0},       // 5
        {"E", "F", 0, 0},

        {"E", "G", "F", 0},

        {"E", "I", "J", 0},

        {"E", "J", 0, 0},       // 4
        {"F", "G", 0, 0},

        {"F", "J", 0, 0},

        {"G", "H", 0, 0},

        {"H", "I", 0, 0},

        {"I", "J", 0, 0},       // 5

    };
    int triangles_count = 0;

    char szTriange[4] = {0};

for (int s1 = 0; s1 &< segments_count; s1++) { for (int s2 = s1 + 1; s2 &< segments_count; s2++) { if (segments[s1].end != segments[s2].start ) { continue; } if (segments[s1].point1 == segments[s2].end || segments[s1].point2 == segments[s2].end || segments[s2].point1 == segments[s1].start || segments[s2].point2 == segments[s1].start ) { // 三點共線 continue; } for (int s3 = 0; s3 &< segments_count; s3++) { if (segments[s1].start == segments[s3].start segments[s2].end == segments[s3].end) { if (segments[s3].point1 != segments[s1].end segments[s3].point2 != segments[s1].end) { ++triangles_count; szTriange[0] = segments[s1].start; szTriange[1] = segments[s1].end; szTriange[2] = segments[s2].end; printf("%s ", szTriange); } } } } } printf("%d ", triangles_count); }

運行結果如下:

36個嗎?

用JS換一種思路寫個試試

輸出：

1,2,3

1,2,4

1,2,5

1,2,6

1,2,7

1,2,8

1,3,4

1,3,5

1,4,5

1,5,6

1,5,7

1,5,9

1,6,7

1,6,9

1,7,8

2,3,6

2,3,8

2,4,7

2,5,6

2,5,7

2,5,10

2,6,7

2,6,8

2,6,10

3,5,6

4,5,7

4,5,9

5,6,7

5,7,9

5,7,10

6,7,8

6,7,9

6,7,10

6,8,10

7,9,10

一共 35個三角形代碼：

& &


&
var dotArr=new Array();

var inLineArr=new Array();
var dot = new Array();

dot[0]	=	[2,3,4,5,6,7,8,9]

dot[1]	=	[1,3,4,5,6,7,8,10]

dot[2]	=	[1,2,4,5,6,8]

dot[3]	=	[1,2,3,5,7,9]

dot[4]	=	[1,2,3,4,6,7,9,10]

dot[5]	=	[1,2,3,5,7,8,9,10]

dot[6]	=	[1,2,4,5,6,8,9,10]

dot[7]	=	[1,2,3,6,7,10]

dot[8]	=	[1,4,5,6,7,10]

dot[9]	=	[2,5,6,7,8,9]
var inLineArray= new Array([1,3,6,8],[1,4,7,9],[2,3,4,5],[2,7,8,10],[5,6,9,10]);
for(i=0;i&2)

{

for (;i&1){

		for (;j&<(inLineArray[l].length);j++){
		var k=j+1
		if(inLineArray[l].length-k&>0){

		for (;k&");

k=k+1;

}
}
document.write("一共 "+k+"個三角形&")
&

&

&

大概意思：

1. 定義二維數組，標出每個點可以連接到的其他點

2. 遍歷所有可能出現的三角形組合（包括三點共線的情況）

3. 遍歷所有共線，去除共線情況，輸出

4. 我寫的仍然很爛

正確答案是35：

#!/usr/bin/python


# total 10 dots

alldots=10

# total 10 lines

allines=10

lines=[[1,2],[2,3],[3,4],[4,5],[1,5],[1,3,7,8],[2,4,8,9],[2,5,6,7],[3,5,9,10],[1,4,6,10]]
def isline(a,b):

    for i in range(0,allines):

        if( (a in lines[i]) and (b in lines[i])) :

            return 1

    return 0

def isline3(a,b,c): for i in range(0,allines): if( (a in lines[i]) and (b in lines[i]) and (c in lines[i])) : return 1 return 0 def istriangle(a,b,c): if( isline(a,b) and isline(a,c) and isline(b,c) ): if(isline3(a,b,c) ) : return 0 else : return 1 return 0 count=0 # for each 3 dots check if they are a triangle. for i in range(1,alldots+1) : for j in range(i+1,alldots+1) : for k in range(j+1,alldots+1) : if ( istriangle(i,j,k) ) : print(i,j,k ) count = count+1 print ("total have ",count," triangles")

這張圖中能數出多少個三角形？

根據前面的 @謝丹的思路，用racket重寫了一遍。答案的確是35。

代碼如下：

#lang racket (require rackunit) (require racket/local) ;; data difinitions ; a Dot is a N represent a dot on the line. ; ; a Line is one of: ; (list Dot Dot) ; (append (list Dot Dot) [List-of Dot]) ; the first two Dot represent the endpoints; ; the other Dot represent the line"s crossing points.


; a Graph is a [List-of Line]

; all lines of the Graph is not the same.
;; constants

(define alldots-1 10) ; total 10 dots

(define lines-1

  "((1 2)

    (2 3)

    (3 4)

    (4 5)

    (5 1)

    (1 3 7 8)

    (3 5 9 10)

    (5 2 6 7)

    (2 4 8 9)

    (4 1 10 6)))
;; functions

; Number Graph -&> N

; produces the number of triangles in the given g.

(define (count-triangles dot-num g)

  (local ((define max-num (add1 dot-num))

          ; [List-of Number] -&> Number

          ; produces the sum of the given lon

          (define (sum lon)

            (foldr + 0 lon))

          ; N -&> [List-of N]

          ; produces a default list of number which starts from

          ; the given s and end at max-num, the step is 1.

          (define (default-range s)

            (range s max-num 1))

          ; [List-of Dot] -&> Boolean

          ; determine whether points are all on the same line of g.

          (define (is-line? points)

            (local (; [List-of Dots] -&> [Line -&> Boolean]

                    ; produces a function which determine whether lop is on the

                    ; line

                    (define (on-line lop)

                      (lambda (line) (andmap (lambda (p) (member p line)) lop)))

                    (define satisfied-lines

                      (filter (on-line points) g))

                    ; Graph -&> Boolean

                    ; determine whether l has a line.

                    (define (exist-line? l)

                      (not (empty? l))))

              (exist-line? satisfied-lines)))

          ; [List Dot Dot Dot]-&> Boolean

          ; determine whether the given lo3p could consist a triangle on g.

          (define (is-triangle? lo3p)

            (let* ([3p (list-&>vector lo3p)]

                   [a (vector-ref 3p 0)]

                   [b (vector-ref 3p 1)]

                   [c (vector-ref 3p 2)])

              (and (andmap (lambda (lop) (is-line? lop))

                           `((,a ,b) (,b ,c) (,c ,a)))

                   (not (is-line? lo3p)))))

          )

    (sum (for*/list ((a (default-range 1))

                     (b (default-range (add1 a)))

                     (c (default-range (add1 b))))

           (if (is-triangle? `(,a ,b ,c)) 1 0)))))

;; test (check-equal? (count-triangles alldots-1 lines-1) 35 "not the correct number of triangles.")

先從由一個三角形開始看有10個再看由2個三角形構成的有10個由3個三角形構成的有10個由四個構成的沒有由5個構成的有5個所以一共有35個

謝邀。居然有人邀請我回答這種問題。。

數了兩遍，我覺得是35個吧。。就分一下類然後數一數。