填空题下面是求s=1+3+5+…+101的伪代码,在“-”上填正确的内容.
s←________????????????
i←________??????????????????
while???i________
i←________
End??while
Print???s.
网友回答
0 1 ≤101 i+2解析分析:先对s赋初始值0,i赋初始值1,然后运行循环语句,判定此时的i是否满足条件,满足条件则执行循环体,因求和一直加到101,可得判定框的条件,因是奇数的和故i←i+2,从而得到结论.解答:s←___0__????????????i←___1_______???????????????????while???i___≤101_s←s+ii←____i+2_____End??whilePrint???s故