去年算法作業(yè)答案第一

limT(N)-T(N)=

3n2+10n-

Nfi

T(N2222

=

= =nfi

3n

nfi￥

nfi￥3n/10n n +2n-2

lim

10 =lim 10

=lim =

10

nfi n1021+1-

+2

nfi￥n10

+2

nfi

10·1+n1

1+

lim =0O(log

nfi

21+n=log 3log =nfi

3logn-3logn=03logn

lim10nlog3-10nlog3=nfi

10nlog 2,logn,n3,20n,4n,3,n! =3·2

3·2n6

n12=64n2n1=)W(log te<classintBinarySearch1(Typea[],constType&x,intn){intleft=0;intright=n-while(left<=intmiddle=if(x==a[middle])returnif(x>a[middle])left=elseright=}return-} te<classintBinarySearch1(Typea[],constType&x,intn){intleft=0;intright=n-while(left<right-intmiddle=if(x<a[middle])right=elseleft=}if(x==a[left])returnreturn-}

2－2 如：序列2345，x=5如：序列2345，x=5故return te<classintBinarySearch1(Typea[],constType&x,intn){intleft=0;intright=n-while(left+1!=intmiddle=if(x>=a[middle])left=elseright=}if(x==a[left])returnreturn-} te<class

2－2(3)中whileleft1!等同于（2）的whileleftright-(3)中ifxa[middle])leftelseright=if(x<a[middle])right=elseleft=intBinarySearch1(Typea[],constType&x,intn){if(n>0&&intleft=0;intright=n-while(left<intmiddle=if(x<a[middle])right=middle-elseleft=}if(x==a[left])return}return-}

如：序列2345，x=55 te<classintBinarySearch1(Typea[],constType&x,intn){if(n>0&&intleft=0;intright=n-while(left<intmiddle=if(x<a[middle])right=middle-elseleft=}if(x==a[left])return}return-} te<classintBinarySearch1(Typea[],constType&x,intn){if(n>0&&intleft=0;intright=n-while(left<intmiddle=if(x<a[middle])right=middle-elseleft=}if(x==a[left])return}return-}

2－2如：序列2345，x=5如：序列22444，x=462－2 te<classintBinarySearch1(Typea[],constType&x,intn){if(n>0&&intleft=0;intright=n-while(left<intmiddle=if(x<a[middle])right=elseleft=}if(x==a[left])return}return-}

如：序列2345，x=27dP(x)=(x-ni)

d

(x-ni

i=1+d

(x-ni)=P1(x)T(d)=

O(1),d=12T(d/2)+O(dlogd),d>1

T(d)=dlogd+dlog(d/2)+dlog(d/4)T(d)=O(dlog2d8

G(n+1)= 1G

T(n)=

1,n=1nT(n-1)+O(n2),n>1 92-2馬的HamiltonT(