/*****************************************************************************/
/* */
/* COUNT PAIRS OF CONSECUTIVE ROOTS OF X**(P-1)=1(MOD P**2) */
/* 02/13/07 (dkc) */
/* */
/* Note: This program is the same as "test1s.c" except that larger p values*/
/* can be processed. */
/* */
/*****************************************************************************/
#include <stdio.h>
#include <math.h>
#include "input.h"
void mul64_64(unsigned int a0, unsigned int a2, unsigned int m0,
unsigned int m2, unsigned int *product);
void div128_64(unsigned int a0, unsigned int a1, unsigned int a2,
unsigned int a3, unsigned int *quotient, unsigned int d2,
unsigned int d3);
void sub128(unsigned int *a, unsigned int *b);
void mul32_32(unsigned int a0, unsigned int m0, unsigned int *product);
void div64_32(unsigned int *dividend, unsigned int *quotient,
unsigned int divisor);
void sub64(unsigned int *a, unsigned int *b);
int main () {
unsigned int g,j,k,j1,h,jh,i,nq,S[4],T[4],U[2],V[2],W[2],X[2];
unsigned int m[50000*2];
int q,f,ilb[50001],irb[50001];
unsigned int s[101];
FILE *Outfp;
Outfp = fopen("output.dat","w");
/**********************/
/* clear histogram */
/**********************/
for (i=0; i<101; i++)
s[i]=0;
/*************************************/
/* load primes and primitive roots */
/*************************************/
for (g=0; g<insize; g++) {
j=input[2*g]; // p (prime)
if (j<60000)
continue;
k=input[2*g+1]; // primitive root
j1=j-1; // p-1
jh=j1/2; // (p-1)/2
mul32_32(j, j, U); // p*p
V[0]=0;
V[1]=1;
sub64(U, V); // p*p-1
div64_32(V, V, 2); // (p*p-1)/2
/**********************************/
/* compute least residue of k**p */
/**********************************/
T[0]=0;
T[1]=k;
for (i=1; i<j; i++) {
mul64_64(T[0], T[1], 0, k, S);
div128_64(S[0], S[1], S[2], S[3], T, U[0], U[1]);
mul64_64(T[2], T[3], U[0], U[1], T);
sub128(S, T);
T[0]=T[2];
T[1]=T[3];
}
m[0]=T[0];
m[1]=T[1];
W[0]=V[0];
W[1]=V[1];
sub64(m, W);
if ((W[0]&0x80000000)==0)
sub64(U, m);
/*****************/
/* compute roots */
/*****************/
for (i=1; i<jh; i++) {
mul64_64(m[2*i-2], m[2*i-1], m[0], m[1], S);
div128_64(S[0], S[1], S[2], S[3], T, U[0], U[1]);
mul64_64(T[2], T[3], U[0], U[1], T);
sub128(S, T);
m[2*i]=T[2];
m[2*i+1]=T[3];
W[0]=V[0];
W[1]=V[1];
sub64(m+2*i, W);
if ((W[0]&0x80000000)==0)
sub64(U, m+2*i);
}
if ((m[2*jh-2]!=0)||(m[2*jh-1]!=1)) {
printf("error: p=%d r=%d %d %d \n",j,k,m[2*jh-2],m[2*jh-1]);
goto zskip;
}
/*****************************************/
/* sort roots using monkey-puzzle sort */
/*****************************************/
nq=0;
ilb[1]=0;
irb[1]=0;
for (h=2; h<=jh; h++) {
ilb[h]=0;
irb[h]=0;
q=1;
lb: W[0]=m[2*q-2];
W[1]=m[2*q-1];
sub64(m+2*(h-1), W);
if ((W[0]&0x80000000)==0)
goto le;
if(ilb[q]==0)
goto ld;
q=ilb[q];
goto lb;
ld: irb[h]=-q;
ilb[q]=(int)h;
continue;
le: if (irb[q]<0)
goto lf;
if (irb[q]==0)
goto lf;
q=irb[q];
goto lb;
lf: irb[h]=irb[q];
irb[q]=(int)h;
}
f=-1;
q=1;
goto li;
lh:q=ilb[q];
li:if (ilb[q]>0)
goto lh;
lj:f=f+1;
if (f!=0) {
if (((m[2*q-2]-X[0])==0)&&((m[2*q-1]-X[1])==1))
nq=nq+1;
}
X[0]=m[2*q-2];
X[1]=m[2*q-1];
if (irb[q]<0)
goto lk;
if (irb[q]==0)
goto ll;
q=irb[q];
goto li;
lk:q=-irb[q];
goto lj;
/*************************************/
/* count pairs of consecutive roots */
/*************************************/
ll:s[nq]=s[nq]+1;
printf(" p=%d, %d \n",j,nq*2);
}
fprintf(Outfp,"\n");
printf("\n");
for (i=0; i<101; i++) {
fprintf(Outfp," i=%d, %d \n",i,s[i]);
printf(" i=%d, %d \n",i,s[i]);
}
zskip:
fclose(Outfp);
return(0);
}