英语翻译Let S = s1 s2...s2n be a well-formed string of parentheses.S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).q By a

来源：学生作业帮助网 编辑：六六作业网 时间：2025/01/25 03:47:43
英语翻译LetS=s1s2...s2nbeawell-formedstringofparentheses.Scanbeencodedintwodifferentways:qByanintegerseq

英语翻译Let S = s1 s2...s2n be a well-formed string of parentheses.S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).q By a
英语翻译
Let S = s1 s2...s2n be a well-formed string of parentheses.S can be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis,say a in S,we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a.(W-sequence).
Following is an example of the above encodings:
S (((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456
请把意思说清楚~
不要谷歌在线翻译的东西~

英语翻译Let S = s1 s2...s2n be a well-formed string of parentheses.S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).q By a
s是一组括弧.有两种编码方法.
方法一,可获得数列P.其中每个Pi是指在第i个右括弧左边的左括弧数目.(((()比如以上在第一个右括弧左边有4个左括弧.所以P1是4,p2是5...
方法二,可获得数列W.把这堆括弧左右一一对应起来.Wi是第i个右括弧是从和它对应的左括弧开始数起的第一个右括弧.对于第一个右括弧,它对应左括弧就是它左面的第一个.所以从这个左括弧数起,它是第一个右括弧.所以w1是1.而第四个右括弧对应的左括弧是最左面第三个括弧,从这个左括弧数起,它的右括弧是第4个右括弧,所以w4是4.

(s3-s2)/s2-(s2-s1)/s1=? c语言语句分析 while(s1[i]&&s2[i]&&s1[i]==s2[i])void main(){char s1[300],s2[300];int result,i=0;printf(请输入两个字符串：)；scanf(%s%s,s1,s2);while(s1[i]&&s2[i]&&s1[i]==s2[i]) i++;result=s1[i]-s2[i];printf(字符串%s,s1);if(resu Δs＝s2-s1=s3-s2=.=aT^2 那s6-s2呢 英语翻译Let S = s1 s2...s2n be a well-formed string of parentheses.S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).q By a 以一直角三角形abc三边为直径的半圆面积分别是S1、S2、S3,直角三角形的面积是S,则S、S1、S2、S3的关系为?(选择） A、S=S1+S2+S3 B、S1=S2+S3 C、S=S1+S2 D、S=S1 while(*s1&&*s2&&*s2++==*s1++);怎样理解 为什么(s1/t1)/(s2/t2)=(s1/s2)/(t1/t2) S2-S1=at^2S1 S2指什么 下面函数的功能是 a(s1,s2) char s1[],s2[];{while(*s2++=*s1++)} 证明：对于任意集合S1和S2,S1并S2=S2并S1 ：对于任意集合S1和S2 请证明：S1∪S2=S2∪S1 △ABC中,DE‖BC,EF‖AB,如果S△ADE=S1,S△EFC=S2,试用S1,S2表示S三△ABC s1=(√1)/2,s2=(√2)/2,s3=(√3)/2,求S1^2+S2^2+S^3+……S^10 一物从斜面顶端静开始匀加速下滑到斜面底端,初3s经路程S1,末3s经路程,S2,S2-s1=1.2,s1:s2=3:7, 物体沿斜面顶端由静止开始做匀加速,最初3s内位移s1,最后3s内位移s2.已知s2-s1=6m,s1:s2=3:7,斜面总长? 物体沿斜面顶端由静止开始做匀加速,最初3s内位移s1,最后3s内位移s2.已知s2+s1=1.2m,s1:s2=3:7,斜面总长? 物体沿斜面顶端由静止开始做匀加速,最初3s内位移s1,最后3s内位移s2.已知s2-s1=1.2m,s1:s2=3:7,斜面总长 物体在斜面顶端由静止匀加速下滑,最初4s内经过的路程为s1,最后4s内经过的路程为s2,且s2-s1=8m,s1:s2=1: