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数学建模

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数学建模. 程 序 设 计. clear clc n=5; for i=1:n for j=1:n A(i,j)=n*(i-1)+j; end end A. clear clc n=5; for i=1:n for j=1:n A(i,j)=n*(i-1)+j; end end A’. (1). (2). clear clc n=5; for i=1:n for j=1:n if mod(i,2)==1 A(i,j)=n*(i-1)+j; else

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数学建模

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  1. 数学建模 程 序 设 计

  2. clear clc n=5; for i=1:n for j=1:n A(i,j)=n*(i-1)+j; end end A clear clc n=5; for i=1:n for j=1:n A(i,j)=n*(i-1)+j; end end A’ (1) (2)

  3. clear clc n=5; for i=1:n for j=1:n if mod(i,2)==1 A(i,j)=n*(i-1)+j; else A(i,j)=n*(i)+1-j; end end end A clear clc n=5; for i=1:n for j=1:n if mod(i,2)==1 A(i,j)=n*(i-1)+j; else A(i,j)=n*(i)+1-j; end end end A' (3) (4)

  4. clear clc n=10;k=1;r=1; for i=2:2*n if i>n+1 r=i-n; end for j=r:i-r A(j,i-j)=k; k=k+1; end end A clear clc k=1; for m=1:5 for i=1:5; for j=1:5 if abs(i-j)==m-1; A(i,j)=k; end end end k=k+1; end A (6) (5)

  5. clear clc k=1; for m=1:5 for i=k:5; for j=k:5 if (i==m | j==m) A(i,j)=k; end end end k=k+1; end A clear clc n=5; r1=4; r2=1; r3=1; r4=5; x=[]; for i=1:5 x=[x,ones(1,r1)*r2,r3:(-1)^(i-1):r4]; r1=r1-1; r2=5-r2+1+(1+(-1)^i)/2;r2 r3=5-r3+1+(1+(-1)^i)/2;r3 r4=5-r4+1+(1+(-1)^(i+1))/2;r4 end x (8) (7)

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