@༺ཌༀ⚡☢.TLE君⚡ༀད༻
> 内存
> 你以为的：
a[0][0] a[0][1] a[0][2] a[1][0] a[1][1] a[1][2] a[2][0] a[2][1] a[2][2]
> 实际上：
a0 a0 a0 a1 a1 a1 …… 0 1 2 0 1 2 ……
> 就能得到：
address=baseAddress+(i∗numbercolums+j)∗sizeof(elementType)address=baseAddress+(i*numbercolums+j)*sizeof(elementType)address=baseAddress+(i∗numbercolums+j)∗sizeof(elementType)
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> 声明
> 输入
1.0:
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2.0:
> SO:
1 0 0 0 2 0 0 0 3 0 0 0
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1 2 3 4 0 0 0 0 0 0 0 0
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> 函数&指针
> 函数声明
上下在={1，2，3，4，5，6}是相同
> 一道平平无奇の选择题
访问A[0][3]会？（多选）
* A.报错
* B.随机值
* C.崩坏……
答案：B,C -……-
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> 指针
*P是一个指针变量,只能存A[0]的地址
A[0+1]是2；
同理*P=1,1+1=21+1=21+1=2
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> SCANF? SCANF!
3存入了A[2]の地址,A={1,2,3}
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> 多维数组の引用
把Aの第1行の4列存入&R
> 来点科技与狠活：
