Find the minimum positive integer x,a^x=1 (mod M) Order template

Order of integers: set A and n are positive integers of the a^x=1, making the smallest positive integer x (mod n) The order of a modulo n

//order Template: A^x=1 (mod M), call Getjie (a,m)//Input: 10^10>a,m>1//output: No solution returned-1, with a solution to return the smallest positive integer x//complexity: O (m^ (0.5))Long LonggcdLong LongALong Longb) {    if(b==0)returnA; returnGCD (b,a%b);}//Euler function: Complexity O (n^ (0.5)), returns the number of all and N-1,n-1 in [Long LongPhiLong Longx) {    Long Longsum=x;  for(Long LongI=2; i*i<=x;i++)    {        if(x%i==0) {sum=sum-sum/i;  while(x%i==0) x/=i; }    }    if(x!=1) sum=sum-sum/x; returnsum;}Long LongMod_mul (Long LongALong LongBLong LongMoD) {    Long Longsum=0;  while(b) {if(b&1) sum= (sum+a)%MoD; b>>=1; A= (a+a)%MoD; }    returnsum;}//A^b%mod Fast PowerLong LongQuk_mul (Long LongALong LongBLong LongMoD) {    Long Longqsum=1;  while(b) {if(b&1) qsum=Mod_mul (QSUM,A,MOD); b>>=1; A=Mod_mul (A, A, Mod); }    returnqsum;}Long LongGetjie (Long LongALong LongM) {    Long LongD =gcd (a,m); if(d! =1)return-1; Long Longm=Phi (M); //and then split the M    Long Longprm[ +],prmcnt[ +]; intPcnt=0; memset (prmcnt,0,sizeof(prmcnt)); Long LongTM =m;  for(Long LongI=2; i*i<=tm;i++)    {        if(tm%i==0) {prm[pcnt]=i;  while(tm%i==0) {prmcnt[pcnt]++; TM/=i; } pcnt++; }    }    if(tm!=1) {prm[pcnt]=TM; PRMCNT[PCNT]=1; Pcnt++; }     for(intI=0; i<pcnt;i++)    {         for(intj=0; j<prmcnt[i];j++)        {            if(Quk_mul (A, M/prm[i], m) = =1) {m/=Prm[i]; }        }    }    returnm;}

Plus a list of the primes. Theoretically it will be 10 times times faster.

//order Template: A^x=1 (mod M), call Getjie (a,m)//Input: 10^10>a,m>1//output: No solution returned-1, with a solution to return the smallest positive integer x//complexity: m^ (0.5)Long Longprime[100100];Long LonggcdLong LongALong Longb) {    if(b==0)returnA; returnGCD (b,a%b);}//Euler function: Complexity O (n^ (0.5)), returns the number of all and N-1,n-1 in [Long LongPhiLong Longx) {    Long Longsum=x; Long Longp = prime[0];  for(intI=0;p *p<=x;i++)    {        if(0= = Xp) {sum=sum-sum/p;  while(x%p==0) x/=p; } P=prime[i+1]; }    if(x!=1) sum=sum-sum/x; returnsum;}Long LongMod_mul (Long LongALong LongBLong LongMoD) {    Long Longsum=0;  while(b) {if(b&1) sum= (sum+a)%MoD; b>>=1; A= (a+a)%MoD; }    returnsum;}//A^b%mod Fast PowerLong LongQuk_mul (Long LongALong LongBLong LongMoD) {    Long Longqsum=1;  while(b) {if(b&1) qsum=Mod_mul (qsum,a,mod);//mod is not a very big time. This step can remove the B>>=1; A=Mod_mul (A, A, Mod);/}returnqsum;}Long LongGetjie (Long LongALong LongM) {    Long LongD =gcd (a,m); if(d! =1)return-1; Long Longm=Phi (M); //and then split the M    Long Longprm[ +],prmcnt[ +]; intPcnt=0; memset (prmcnt,0,sizeof(prmcnt)); Long LongTM =m; Long Longp=prime[0];  for(Long LongI=0;p *p<=tm;i++)    {        if(tm%p==0) {prm[pcnt]=p;  while(tm%p==0) {prmcnt[pcnt]++; TM/=p; } pcnt++; } P=prime[i+1]; }    if(tm!=1) {prm[pcnt]=TM; PRMCNT[PCNT]=1; Pcnt++; }     for(intI=0; i<pcnt;i++)    {         for(intj=0; j<prmcnt[i];j++)        {            if(Quk_mul (A, M/prm[i], m) = =1) {m/=Prm[i]; }        }    }    returnm;}voidGetprime (Long LongUp ) {    BOOLpmark[200100]; memset (Pmark,0,sizeof(Pmark)); intPcnt=0; Prime[pcnt++]=2;  for(intI=2; i<=up;i+=2) pmark[i]=1;  for(intI=3; i<=up;i+=2)    {        if(pmark[i]==0) {prime[pcnt++ ] =i;  for(intj=i+i;j<=up;j+=i) {pmark[j]=1; }        }    }}

Find the minimum positive integer x,a^x=1 (mod M) Order template