2014-08-28 102 views
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我正在嘗試確定三重嵌套循環的頻率計數。如何計算三重嵌套循環中的頻率計數

for i = 1 to n do 
    for j = 1 to i do 
     for k = i to j do 
      x = x + 1 

我知道聲明x = x + 1將不會執行,直到我實現n

關於如何開始任何提示/建議的價值?

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這個問題會更適合http://math.stackexchange.com。 – 2014-08-28 03:15:00

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「x = x + 1將不會得到執行,直到我達到n的值」不是真的。當i = 1和j = 1且k = 1時,它將被執行。然後再次當i = 1和j = 1和k = 2時,on和on ... – jch 2014-08-28 03:16:51

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對於每個'i','x = x + 1'執行'r(r + 1)/ 2'次其中'r =絕對(1-i)+ 1' – hk6279 2014-08-28 03:40:02

回答

1

我們以4和5爲例。當i = 4,

... 
    for j = 1 to 4 do 
     for k = 4 to j do 
      x = x + 1 

...j = 1 
    for k = 4 to 1 do // 4 times 
     x = x + 1 
...j = 2 
    for k = 4 to 2 do // 3 times 
     x = x + 1 
...j = 3 
    for k = 4 to 3 do // twice 
     x = x + 1 
...j = 4 
    for k = 4 to 4 do // once 
     x = x + 1 

當i = 5,

... 
    for j = 1 to 5 do 
     for k = 5 to j do 
      x = x + 1 

...j = 1 
    for k = 5 to 1 do // 5 times 
     x = x + 1 
...j = 2 
    for k = 5 to 2 do // 4 times 
     x = x + 1 
...j = 3 
    for k = 5 to 3 do // 3 times 
     x = x + 1 
...j = 4 
    for k = 5 to 4 do // twice 
     x = x + 1 
...j = 5 
    for k = 5 to 5 do // once 
     x = x + 1 

圖案?

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謝謝!我能弄明白了! – 2014-09-02 18:45:09