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@@ -0,0 +1,10 @@
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c = input('请输入一个字符:', 's');
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if c >= 'A' && c <= 'Z'
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disp(char(abs(c) + abs('a') - abs('A')));
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elseif c >= 'a' && c <= 'z'
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disp(char(abs(c) - abs('a') + abs('A')));
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elseif c >= '0' && c <= '9'
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disp(abs(c) - abs('0'));
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else
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disp(c);
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end
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@@ -0,0 +1,6 @@
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mysum=0,i=1;
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while(i<=100)
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mysum=mysum+i
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i=i+1 ;
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end
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mysum
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@@ -0,0 +1,13 @@
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A=[1 2 3;4 5 6;7 8 9];
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[r,c]=size(A);
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% 判断并把A中大于8和小于2的元素全部调整为0
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for i=1:1:r
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for j=1:1:c
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if (A(i,j)>8 || A(i,j)<2)
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A(i,j)=0;
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end
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end
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end
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% 改写
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b=(A <= 8 & A >= 2); % 判断A中小于等于8和大于等于2的元素,结果是逻辑值0/1
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A=A.*b %实现A中大于8和小于2的元素全部调整为0,基于矩阵的点乘
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