Bzoj 2242: [SDOI2011] Calculator & BSGS Algorithm notes

The main difficulty of this question is how to solve the third question.

So we need to know what kind of a way bsgs is.

First bsgs is a kind of meet in the middle (poke below)

http://m.blog.csdn.net/blog/zentropy/11200099

Read the link and see the following:

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For a prime number p we know by the Fermat theorem that y^xmodp up to (p-1) times is a cyclic section

So if there's a solution, x must be 0 to p-1.

So we just need to know if X takes 0 to p-1 if there's a solution.

According to the thought of meet in the middle M=sqrt (p-1)

Then we just need to find out 0~m-1 (if there is a solution to exit here)

And then find out m,2m,3m......nm (nm<=p-1)

Ask y^ (0~m-1) If there are and y^km multiply MODP equals Z.

It is clear, however, that this operation is done directly by sqrt (n) *sqrt (n) =n

So I learned how to find the inverse of the y^km and then we can multiply the inverse of the modp meaning with Z

Then ask y^ (0~m-1) If there is no equal to it

Doing this is sqrt (n) the complexity of *hash

With the manual hashing technique, the hash complexity can be considered as 1.

More lazy words directly with map to hash is log (n)

#include <bits/stdc++.h>using namespaceStd;map<int,int>MP;intSolve1 (intXintYintMoD) {    Long LongT=x,re=1;  while(y) {if(y&1) Re=re*t%MoD; T=t*t%MoD; Y>>=1; }    return(int) re;}intEXGCD (intAintBint&x,int&y) {    if(!b) {x=1; Y=0; returnA; }    intt,d; D=EXGCD (b,a%b,x,y); T=x; X=y; Y=t-(A/b) *x; returnD; }voidSolve2 (intYintZintp) {    intX,yy; intD=EXGCD (Y,P,X,YY); if(z%d) {puts ("Orz, I cannot find x!"); return; } x=(Long Long) x* (z/d)%p; X= (x<0? x+p:x); printf ("%d\n", x);}BOOLSolve3 (intYintZintp) {//if (z>=p)//return 0;y%=p; if(!y) {if(z)return 0; Puts ("1"); return 1;    } mp.clear (); intM=ceil (sqrt (P-1)); Long Longt=1;  for(intI=0; i<m;++i) {if(t==z) {printf ("%d\n", i); return 1; }        if(!mp[t]) mp[t]=i+1; Else            return 0; T=t*y%p; }    intInv=solve1 (y,p-1-m,p); T=Z;  for(inti=m;i<=p-2; i+=m) {T=t*inv%p; if(Mp[t]) {printf ("%d\n", i+mp[t]-1); return 1; }    }    return 0;}intMain () {intt,ca,y,z,p; scanf ("%d%d",&t,&CA);  while(t--) {scanf ("%d%d%d",&y,&z,&p); if(ca==1) printf ("%d\n", Solve1 (y,z,p)); Else if(ca==2) solve2 (y,z,p); Else if(ca==3)            if(!Solve3 (y,z,p)) puts ("Orz, I cannot find x!"); }    return 0;}

Bzoj 2242: [SDOI2011] Calculator & BSGS Algorithm notes