dimension a (3) s=0 for k=1 to 3 a(k)=3-k s=s+k endfor a(1),a(2),a(3),s

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dimensiona(3)s=0fork=1to3a(k)=3-ks=s+kendfora(1),a(2),a(3),sdimensiona(3)s=0fork=1to3a(k)=3-ks=s+ken

dimension a (3) s=0 for k=1 to 3 a(k)=3-k s=s+k endfor a(1),a(2),a(3),s
dimension a (3) s=0 for k=1 to 3 a(k)=3-k s=s+k endfor a(1),a(2),a(3),s

dimension a (3) s=0 for k=1 to 3 a(k)=3-k s=s+k endfor a(1),a(2),a(3),s
so easy .
first ,we move on to k.If k equals 1,then we have a(1)=3-1=2.Thus S=0+k=0+1=1
next,again,we continue this process ,which we have a(2)=3-2=1.And S=1+2=3.
Finally,if k is 3,we will get a(3)=3-3=0 ,and s=3+3=6.
Therefore,a(1)=2,a(2)=1,a(3)=0 and s=6.
楼主下次发题请用中文.请用自然语言& 别写循环语句.

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