📄 螺旋方阵--fevernova的博客.htm
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style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes"> </SPAN>printf("%4d",t)；
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style="mso-spacerun: yes"> </SPAN>}
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lang=EN-US
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style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes"> </SPAN>printf("\n")；
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style="mso-spacerun: yes"> </SPAN><SPAN
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style="mso-spacerun: yes"> </SPAN>}
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lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体">}
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<P class=MsoNormal style="tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
lang=EN-US style="FONT-SIZE: 12pt; FONT-FAMILY: 宋体">
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<P class=MsoNormal style="tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
style="FONT-SIZE: 12pt; COLOR: blue; FONT-FAMILY: 宋体">第<SPAN
lang=EN-US>5种解法：递归方法。
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style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体">分析：根据本题图形的特点，我们可以构造一个递归算法。我们可以将边长为<SPAN
lang=EN-US>N的图形分为两部分：第一部分最外层的框架，第二部分为中间的边长为N-2的图形。
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style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体">对于边长为<SPAN
lang=EN-US>N的正方型，若其中每个元素的行号为i（1≤i≤N），列号为j（1≤j≤N），第1行第1列元素表示为a<SUB>1,1</SUB>（a11=1），则有：
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lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes"> </SPAN>对于最外层的框架可以用以下数学模型描述：
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style="MARGIN-LEFT: 30.85pt; TEXT-INDENT: -30.85pt; mso-char-indent-count: -2.57; mso-char-indent-size: 12pt; tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes">
</SPAN>上边：<SPAN style="mso-spacerun: yes">
</SPAN>a<SUB>1,j</SUB>=a<SUB>1,1</SUB>+j-1<SPAN
style="mso-spacerun: yes">
</SPAN>(j≠1)
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lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
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</SPAN>右边：<SPAN style="mso-spacerun: yes">
</SPAN>a<SUB>i,N</SUB>=a<SUB>1,1</SUB>+N+i-2<SPAN
style="mso-spacerun: yes">
</SPAN>(i≠1)
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lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes">
</SPAN>下边：<SPAN style="mso-spacerun: yes">
</SPAN>a<SUB>i,1</SUB>=a<SUB>1,1</SUB>+4N-i-3<SPAN
style="mso-spacerun: yes"> </SPAN>(i≠1)
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style="MARGIN-LEFT: 30.85pt; TEXT-INDENT: -30.85pt; mso-char-indent-count: -2.57; mso-char-indent-size: 12pt; tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes">
</SPAN>左边：<SPAN style="mso-spacerun: yes">
</SPAN>a<SUB>N,j</SUB>=a<SUB>1,1</SUB>+3N-2-j<SPAN
style="mso-spacerun: yes"> </SPAN>(j≠1)
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<P class=MsoPlainText
style="MARGIN-LEFT: 30.85pt; TEXT-INDENT: -30.85pt; mso-char-indent-count: -2.57; mso-char-indent-size: 12pt; tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes"> </SPAN>对于内层的边长为N-2的图形可以用以下数学模型描述：
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style="MARGIN-LEFT: 30.85pt; TEXT-INDENT: -30.85pt; mso-char-indent-count: -2.57; mso-char-indent-size: 12pt; tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
style="mso-spacerun: yes"> </SPAN><SPAN
style="mso-spacerun: yes">
</SPAN>左上角元素：a<SUB>i,i</SUB>=a<SUB>i-1,i-1</SUB>+4(N-2i-1)<SPAN
style="mso-spacerun: yes"> </SPAN>(i＞1)
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style="TEXT-INDENT: 24pt; mso-char-indent-count: 2.0; mso-char-indent-size: 12pt; tab-stops: 595.4pt 641.2pt 687.0pt 732.8pt"><SPAN
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体">若令：<SPAN
lang=EN-US>a<SUB>i,j</SUB>=fun(a<SUB>i-1,i-1</SUB>+4(N-2i-1)，当：i＜(N+1)/2且j＜(N+1)/2时，min=MIN(i,j)，则有：
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lang=EN-US
style="FONT-SIZE: 12pt; mso-bidi-font-size: 10.0pt; mso-hansi-font-family: 宋体"><SPAN
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