以下程式會算出 C(N, M)，即從 N 個物品中選出 M 個物品的方法數量。如果 count 的值原先為 0 ，請問計算 C(5, 3) 後，count 的值為何？ unsigned int count = 0; unsigned int getC(unsigned int N, unsigned int M){ count++; if (N == 0) return (N == M ? 1 : 0); else if (M == 0) return 1; else return getC(N-1, M) + getC(N-1, M-1); }
(A) 5
(B) 15
(C) 51
(D) 63

答案：登入後查看
統計： A(7), B(33), C(51), D(20), E(1) #91985

詳解 (共 5 筆)

#389688
不好意思，答案應該有誤，應該是51次~
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#1301917
01=>Text1a(3,5)=a(4,3)a(4,2))
02=>a(3,4)=a(3,3)a(3,2))
03=>a(3,3)=a(2,3)a(2,2))
04=>a(3,2)=a(1,3)a(1,2))
05=>a(3,1)=a(0,3)a(0,2))
06=>a(3,0)=0
07=>a(2,0)=0
08=>a(2,1)=a(0,2)a(0,1))
09=>a(2,0)=0
10=>a(1,0)=0
11=>a(2,2)=a(1,2)a(1,1))
12=>a(2,1)=a(0,2)a(0,1))
13=>a(2,0)=0
14=>a(1,0)=0
15=>a(1,1)=a(0,1)a(0,0))
16=>a(1,0)=0
17=>a(0,0)=1
18=>a(2,3)=a(2,2)a(2,1))
19=>a(2,2)=a(1,2)a(1,1))
20=>a(2,1)=a(0,2)a(0,1))
21=>a(2,0)=0
22=>a(1,0)=0
23=>a(1,1)=a(0,1)a(0,0))
24=>a(1,0)=0
25=>a(0,0)=1
26=>a(1,2)=a(1,1)a(1,0))
27=>a(1,1)=a(0,1)a(0,0))
28=>a(1,0)=0
29=>a(0,0)=1
30=>a(0,1)=1
31=>a(2,4)=a(3,2)a(3,1))
32=>a(2,3)=a(2,2)a(2,1))
33=>a(2,2)=a(1,2)a(1,1))
34=>a(2,1)=a(0,2)a(0,1))
35=>a(2,0)=0
36=>a(1,0)=0
37=>a(1,1)=a(0,1)a(0,0))
38=>a(1,0)=0
39=>a(0,0)=1
40=>a(1,2)=a(1,1)a(1,0))
41=>a(1,1)=a(0,1)a(0,0))
42=>a(1,0)=0
43=>a(0,0)=1
44=>a(0,1)=1
45=>a(1,3)=a(2,1)a(2,0))
46=>a(1,2)=a(1,1)a(1,0))
47=>a(1,1)=a(0,1)a(0,0))
48=>a(1,0)=0
49=>a(0,0)=1
50=>a(0,1)=1
51=>a(0,2)=1

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#413341
原本答案為D,修改為C
(共 13 字，隱藏中）
前往觀看
0
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#363210
請問怎麼解呢
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#1154539
有人可以幫忙詳解ㄇ
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