如何用Lua一维数组存储n阶方阵,计算正反对角线数之和?

2026-04-01 19:131阅读0评论SEO资源
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本文共计132个文字,预计阅读时间需要1分钟。

如何用Lua一维数组存储n阶方阵,计算正反对角线数之和?

pythonarr=[11, 2, 4, 4, 5, 6, 10, 8, -12]def diagonalDifference(arr): dimension=math.sqrt(len(arr)) return abs(arr[int(dimension)-1] - arr[int(dimension)-int(dimension)-1])

如何用Lua一维数组存储n阶方阵,计算正反对角线数之和?

arr = {11, 2, 4, 4, 5, 6, 10, 8, -12} function diagonalDifference(arr) dimesion = math.sqrt(#arr) arr1 = {} sum1 = 0 arr2 = {} sum2 = 0 for i=1,dimesion do for j=1,dimesion do if(i == j)then arr1[j] = arr[1 + (j - 1) * (dimesion + 1)] sum1 = sum1 + arr1[j] --print(arr1[j]) end end end for i=1,dimesion do for j=1,dimesion do if((i + j) == (dimesion + 1))then arr1[j] = arr[dimesion + (j - 1) * (dimesion - 1)] sum2 = sum2 + arr1[j] --print(arr1[j]) end end end return math.abs(sum2 - sum1) end print(diagonalDifference(arr))

本文共计132个文字,预计阅读时间需要1分钟。

如何用Lua一维数组存储n阶方阵,计算正反对角线数之和?

pythonarr=[11, 2, 4, 4, 5, 6, 10, 8, -12]def diagonalDifference(arr): dimension=math.sqrt(len(arr)) return abs(arr[int(dimension)-1] - arr[int(dimension)-int(dimension)-1])

如何用Lua一维数组存储n阶方阵,计算正反对角线数之和?

arr = {11, 2, 4, 4, 5, 6, 10, 8, -12} function diagonalDifference(arr) dimesion = math.sqrt(#arr) arr1 = {} sum1 = 0 arr2 = {} sum2 = 0 for i=1,dimesion do for j=1,dimesion do if(i == j)then arr1[j] = arr[1 + (j - 1) * (dimesion + 1)] sum1 = sum1 + arr1[j] --print(arr1[j]) end end end for i=1,dimesion do for j=1,dimesion do if((i + j) == (dimesion + 1))then arr1[j] = arr[dimesion + (j - 1) * (dimesion - 1)] sum2 = sum2 + arr1[j] --print(arr1[j]) end end end return math.abs(sum2 - sum1) end print(diagonalDifference(arr))