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Co-authored-by: sigonasr2 <sigonasr2@gmail.com>
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src/main.c
217
src/main.c
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#include <stdio.h>
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#include "utils.h"
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#include <stdlib.h>
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/*
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A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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Euler discovered the remarkable quadratic formula:
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1/2 = 0.5
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1/3 = 0.(3)
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1/4 = 0.25
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1/5 = 0.2
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1/6 = 0.1(6)
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1/7 = 0.(142857)
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1/8 = 0.125
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1/9 = 0.(1)
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1/10 = 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
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Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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It turns out that the formula will produce 40 primes for the consecutive integer values . However, when is divisible by 41, and certainly when is clearly divisible by 41.
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https://projecteuler.net/problem=26
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The incredible formula was discovered, which produces 80 primes for the consecutive values . The product of the coefficients, −79 and 1601, is −126479.
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Considering quadratics of the form:
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, where and
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where is the modulus/absolute value of
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e.g. and
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Find the product of the coefficients, and , for the quadratic expression that produces the maximum number of primes for consecutive values of , starting with .
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https://projecteuler.net/problem=27
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*/
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boolean isPrime(int*primeList,int primeListSize,int numb) {
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for (int i=0;i<primeListSize;i++) {
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if (numb==primeList[i]) {
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return true;
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}
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}
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return false;
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}
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int*getFactors(int numb) {
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int*factorList=malloc(sizeof(int)*1);
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int factorListSize=2;
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factorList[0]=1;
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factorList[1]=numb;
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int max=numb;
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for (int i=2;i<max;i++) {
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if (numb%i==0) {
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factorList=realloc(factorList,sizeof(int)*++factorListSize);
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factorList[factorListSize-1]=i;
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if (numb/i!=i) {
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factorList=realloc(factorList,sizeof(int)*++factorListSize);
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factorList[factorListSize-1]=numb/i;
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max=numb/i;
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}
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}
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}
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factorList=realloc(factorList,sizeof(int)*++factorListSize);
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factorList[factorListSize-1]=0;
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return factorList;
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}
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boolean isFactor(int*factorList,int numb) {
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int counter=0;
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while (factorList[counter]!=0) {
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if (numb==factorList[counter]) {
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return true;
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}
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counter++;
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}
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return false;
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}
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const int TARGET_REPEATS_REQUIRED=100000;
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int main(int argc,char**argv) {
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int divider=0;
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int divisor=2;
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int sequence[TARGET_REPEATS_REQUIRED]; //Let's assume for the sake of sanity that 100000 repeating digits is enough. I sure hope so.
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int sequenceLength=0;
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int sequenceRepeat=0;
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int sequenceMarker=-1;
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int longestCycleLength=0;
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int longestCycleDivisor=0;
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int longestSequence[TARGET_REPEATS_REQUIRED/10];
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int*primeList=malloc(sizeof(int)*0);
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int primeListSize=0;
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FILE*f = fopen("archives/primegenerator/primes","r");
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while (fgetc(f)!='{');
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int startMarker=0;
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while (fgetc(f)!='{') {
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startMarker++;
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}
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fseek(f,startMarker,SEEK_SET);
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int n=0;
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while (true) {
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char c;
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int digit=0;
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while ((c=fgetc(f))!=',') {
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digit*=10;
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digit+=c-'0';
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}
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if (digit>=1000) {
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break;
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} else {
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primeList=realloc(primeList,sizeof(int)*++primeListSize);
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primeList[primeListSize-1]=digit;
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for (int a=1;a<1000;a++) {
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for (int b=1;b<=1000;b++) {
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}
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}
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}
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while (divisor<1000) {
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divider=10; //We always start with 10.
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sequenceLength=0;
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sequenceRepeat=0;
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for (int i=0;i<sequenceLength;i++) {
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sequence[i]=0;
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}
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sequenceMarker=-1;
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while (sequenceLength<TARGET_REPEATS_REQUIRED) {
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int result=divider/divisor;
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//printf("%d/%d = %d\n",divider,divisor,result);
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divider=divider%divisor;
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sequence[sequenceLength++]=result;
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divider*=10;
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if (divider==0) {
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//We're solved.
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break;
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}
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}
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/*printf("%d:",divisor);
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for (int i=0;i<sequenceLength;i++) {
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printf("%d",sequence[i]);
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}
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printf("\n");*/
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//We need to look at all combinations of possible repeating sequences.
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//Starting from iteration loop of length 1, and an offset incrementing by 1, see if anything remains the same.
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//Then do this for iteration loop of length 2, offset incrementing by 1, etc.
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//Find it repeating at least 20 times to assume it repeats infinitely.
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boolean isP = isPrime(primeList,primeListSize,divisor);
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int*factors = getFactors(divisor-1);
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int counter=0;
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/*if (isP) {
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printf("Possible length factors of (%d-1):",divisor);
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while (factors[counter]!=0) {
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printf("%d,",factors[counter]);
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counter++;
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}
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printf("\n");
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}*/
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boolean matched=false;
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if (sequenceLength==TARGET_REPEATS_REQUIRED) {
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for (int checkLength=0;checkLength<TARGET_REPEATS_REQUIRED/10;checkLength++) {
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if (isP&&!isFactor(factors,checkLength+1)) {continue;}
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for (int offset=0;offset<sequenceLength+1-checkLength;offset++) {
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int sequenceStore[checkLength+1];
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int requiredCheckAmt=TARGET_REPEATS_REQUIRED/10-2;
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for (int i=0;i<checkLength+1;i++) {
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sequenceStore[i]=sequence[i+offset];
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/*printf("%d:",divisor);
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for (int i=0;i<checkLength+1;i++) {
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printf("%d",sequenceStore[i]);
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}
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0010030090270812437311935807422266800401203610832497492477432296890672016048144433299899699097291875626880641925777331995987963891675025075225677031093279839518555667
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0010030090270812437311935807422266800401203610832497492477432296890672016048144433299899699097291875626880641925777331995987963891675025075225677031093279839518555667
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printf("\n");*/
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}
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//Now if this repeating sequence can be found requiredCheckAmt times, we're golden.
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boolean allMatching=true;
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int currentOffset=0;
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int found=0;
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while (requiredCheckAmt>0&&found<20) {
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for (int i=0;i<checkLength+1;i++) {
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//printf(" Compare %d to %d\n",sequence[offset+currentOffset+checkLength+1+i],sequenceStore[i]);
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if (sequence[offset+currentOffset+checkLength+1+i]!=sequenceStore[i]) {
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allMatching=false;
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goto check;
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}
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}
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requiredCheckAmt--;
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currentOffset+=checkLength+1;
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found++;
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//printf("Found a repeat.");
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}
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check:
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if (allMatching) {
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printf("Longest repeating sequence for %d is of length %d: ",divisor,checkLength+1);
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for (int i=0;i<checkLength+1;i++) {
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printf("%d",sequenceStore[i]);
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}
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printf("\n");
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if (longestCycleLength<checkLength+1) {
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longestCycleLength=checkLength+1;
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longestCycleDivisor=divisor;
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for (int i=0;i<longestCycleLength;i++) {
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longestSequence[i]=sequenceStore[i];
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}
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}
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matched=true;
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goto next;
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}
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}
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}
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} else {
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matched=true;
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printf("Longest repeating sequence for %d is of length 0.\n",divisor);
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}
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next:
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free(factors);
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if (!matched) {
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printf("Not enough data to calculate longest repeating sequence for %d!\nQuitting...\n",divisor);
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return 1;
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}
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divisor++;
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}
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printf("\n\nThe largest repeating sequence length is from %d with %d in length.",longestCycleDivisor,longestCycleLength);
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printf("\n\tSequence: ");
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for (int i=0;i<longestCycleLength;i++) {
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printf("%d",longestSequence[i]);
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}
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printf("\n");
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return 0;
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}
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