IT討論區(83) 點解廢人可以做得比你少,人工比你多,公平咩
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悶死 日日對住個mon
Dev先最快樂 砌吓砌吓咗 楽しい
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有冇巴打做過ibank data 嗰邊既野
想問下做ibank Data analysis 前景
想問下做ibank Data analysis 前景
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做曲狗又燒腦又要日日比人撚
長期對著個mon自閉
fg唔好哂時間做啦
長期對著個mon自閉
fg唔好哂時間做啦
快樂砌曲 毒撚嘅朗癲
做左幾年support狗
想向專少少發展
想知Oracle Database Administration Certified Professional 考完有冇用?
心思思想報systematics
本人淨係得個degree同uipath RPA既cert
唔知點開始好
想向專少少發展
想知Oracle Database Administration Certified Professional 考完有冇用?
心思思想報systematics
本人淨係得個degree同uipath RPA既cert
唔知點開始好
學左新既黑魔法
覺得自己algo真係好差
The Fortran I compiler would expand each operator with a sequence of parentheses. In a simplified form of the algorithm, it would
replace + and – with ))+(( and ))-((, respectively;
replace * and / with )*( and )/(, respectively;
add (( at the beginning of each expression and after each left parenthesis in the original expression; and
add )) at the end of the expression and before each right parenthesis in the original expression.
Although not obvious, the algorithm was correct, and, in the words of Knuth, “The resulting formula is properly parenthesized, believe it or not.”
覺得自己algo真係好差The Fortran I compiler would expand each operator with a sequence of parentheses. In a simplified form of the algorithm, it would
replace + and – with ))+(( and ))-((, respectively;
replace * and / with )*( and )/(, respectively;
add (( at the beginning of each expression and after each left parenthesis in the original expression; and
add )) at the end of the expression and before each right parenthesis in the original expression.
Although not obvious, the algorithm was correct, and, in the words of Knuth, “The resulting formula is properly parenthesized, believe it or not.”
Yes, you can generalize it to + - * / and ^.
+ -> )))+((( ditto for -
* -> ))*((
^ -> )^(
enclose the whole expression with ((( and )))
If you have n different precedence level, use one level of () for the operators with the highest precedence and add one () for each level going down.
+ -> )))+((( ditto for -
* -> ))*((
^ -> )^(
enclose the whole expression with ((( and )))
If you have n different precedence level, use one level of () for the operators with the highest precedence and add one () for each level going down.
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