"POJ" 3243 Clever Y

http://poj.org/problem?id=3243

Test instructions: $a^y \equiv b \pmod{p}$ the smallest $y$. (0<=x, y, p<=10^9)

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include < Iostream>typedef Long Long ll;using namespace Std;int gcd (int a, int b) {return b?gcd (b, a%b): A;} void Exgcd (ll A, ll B, LL &d, LL &x, ll &y) {if (!b) {d=a; x=1; y=0; return;} exgcd (b, a%b, D, y, x); y-=a/b *x; }int ni (int a, int b) {static ll x, Y, D;EXGCD (A, B, D, X, y); return (x+b)%b;} int Ipow (int a, int b, int c) {int x=1; for (; b; b>>=1, a= (LL) a*a%c) if (b&1) x= (ll) x*a%c; return x;} struct H {static const int md=3999997;bool VIS[MD]; int DT[MD], FOO[MD], S[MD], top;void CLR () {while (top) {int x=s[top- -]; Vis[x]=0, Dt[x]=-1; }}void Add (int a, int b) {int x=a%md;while (1) {if (!vis[x] | | dt[x]==a) {if (!vis[x]) s[++top]=x; vis[x]=1; dt[x]=a; foo[ X]=b; Break } ++x; if (X==MD) x=0; }}int find (int a) {int x=a%md;while (1) {if (!vis[x]) return-1, if (dt[x]==a) return foo[x]; ++x; if (X==MD) x=0;}} H;int A, B, Mo;bool SPJ () {if (b==1) puts ("0"); else if (!a &&!b) puts ("1"), else if (!b) puts ("No solution"); else return 0;return 1;} void work () {b%=mo;if (SPJ ()) return;for (int i=0, t=1; i<30; ++i, t= (LL) t*a%mo) if (t==b) {printf ("%d\n", I); return;} int a1=0, d=1, G;while ((G=GCD (A, MO))!=1) {if (b%g) {puts ("No solution"); return; ++a1, Mo/=g, B/=g, d= (LL) A/g*d%mo;} D=ni (d, MO); int m=sqrt (0.5+MO); h.clr (); for (int i=0, t= (LL) B*d%mo; i<m; ++i, t= (LL) t*a%mo) H.add (t, I); for (int i=0, wn= Ipow (A, M, MO), t=1; i<=m; ++i, t= (LL) t*wn%mo) {int x=h.find (t), if (x!=-1) {printf ("%d\n", a1+i*m-x); return;}} Puts ("No solution");} int main () {while (scanf ("%d%d%d", &a, &mo, &b) && (A|B|MO)) work (); return 0;}

　　

The big step of expansion = =

Kneel on your knees: http://hi.baidu.com/aekdycoin/item/236937318413c680c2cf29d4

Since $c$ are not necessarily prime numbers, they cannot be used with the original BSGS algorithm

And that's a solution, but it may not be the only thing.

But it is still easy to think of the enumeration $i$ solve this equation $a^{im} x \equiv b \pmod{p}$ then find $x$ is $a^t$, then the answer is $im+t$.

However, it is found that the size of the solution set is $ (A^{im}, p) $, it may be large = cannot withstand ....

Let's simplify the problem by turning the equation into an equivalent form:

We divide the equation by $d= (A, p) $ and get: $\frac{a}{d} a^m \equiv \frac{b}{d} \pmod{\frac{p}{d}}$. (Of course, if $b$ can't divide $d$, then the equation has no solution spicy = = will $b = xxx$ to be able to know spicy =)

has been $t$ times until $ (a, p_{t}) = 1$, so $d = A^t \frac{1}{\prod_{i=1}^{t}} d_{t}$, then obviously $ (D, p_{t}) = 1$ is not spicy = =

Then get the equivalent form of the original equation $a^h \equiv b_{t} d^{-1} \pmod{p_t}$, solve $h$ then the original question answer is $h+t$ spicy = =

So naked $bsgs$ spicy

But here to pay attention to Oh, there may be $y<t$ situation yo, because $t$ pine upper bound for $log_2 p$, we first enumerate the judgment on the line spicy ....

"POJ" 3243 Clever Y