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Co-authored-by: sigonasr2 <sigonasr2@gmail.com>
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sigonasr2, Sig, Sigo 2022-07-21 18:26:35 +00:00
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commit f0d9bbfaeb
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src/main.c
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@ -1,211 +1,40 @@
#include <stdio.h> #include <stdio.h>
#include "utils.h" #include "utils.h"
#include <stdlib.h>
/* /*
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given: Euler discovered the remarkable quadratic formula:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
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.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part. 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.
https://projecteuler.net/problem=26 The incredible formula was discovered, which produces 80 primes for the consecutive values . The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
, where and
where is the modulus/absolute value of
e.g. and
Find the product of the coefficients, and , for the quadratic expression that produces the maximum number of primes for consecutive values of , starting with .
https://projecteuler.net/problem=27
*/ */
boolean isPrime(int*primeList,int primeListSize,int numb) {
for (int i=0;i<primeListSize;i++) {
if (numb==primeList[i]) {
return true;
}
}
return false;
}
int*getFactors(int numb) {
int*factorList=malloc(sizeof(int)*1);
int factorListSize=2;
factorList[0]=1;
factorList[1]=numb;
int max=numb;
for (int i=2;i<max;i++) {
if (numb%i==0) {
factorList=realloc(factorList,sizeof(int)*++factorListSize);
factorList[factorListSize-1]=i;
if (numb/i!=i) {
factorList=realloc(factorList,sizeof(int)*++factorListSize);
factorList[factorListSize-1]=numb/i;
max=numb/i;
}
}
}
factorList=realloc(factorList,sizeof(int)*++factorListSize);
factorList[factorListSize-1]=0;
return factorList;
}
boolean isFactor(int*factorList,int numb) {
int counter=0;
while (factorList[counter]!=0) {
if (numb==factorList[counter]) {
return true;
}
counter++;
}
return false;
}
const int TARGET_REPEATS_REQUIRED=100000;
int main(int argc,char**argv) { int main(int argc,char**argv) {
int divider=0;
int divisor=2;
int sequence[TARGET_REPEATS_REQUIRED]; //Let's assume for the sake of sanity that 100000 repeating digits is enough. I sure hope so.
int sequenceLength=0;
int sequenceRepeat=0;
int sequenceMarker=-1;
int longestCycleLength=0;
int longestCycleDivisor=0;
int longestSequence[TARGET_REPEATS_REQUIRED/10];
int*primeList=malloc(sizeof(int)*0);
int primeListSize=0;
FILE*f = fopen("archives/primegenerator/primes","r"); FILE*f = fopen("archives/primegenerator/primes","r");
while (fgetc(f)!='{'); int startMarker=0;
while (fgetc(f)!='{') {
startMarker++;
}
fseek(f,startMarker,SEEK_SET);
int n=0;
while (true) { while (true) {
char c; for (int a=1;a<1000;a++) {
int digit=0; for (int b=1;b<=1000;b++) {
while ((c=fgetc(f))!=',') {
digit*=10;
digit+=c-'0';
}
if (digit>=1000) {
break;
} else {
primeList=realloc(primeList,sizeof(int)*++primeListSize);
primeList[primeListSize-1]=digit;
}
}
while (divisor<1000) {
divider=10; //We always start with 10.
sequenceLength=0;
sequenceRepeat=0;
for (int i=0;i<sequenceLength;i++) {
sequence[i]=0;
}
sequenceMarker=-1;
while (sequenceLength<TARGET_REPEATS_REQUIRED) {
int result=divider/divisor;
//printf("%d/%d = %d\n",divider,divisor,result);
divider=divider%divisor;
sequence[sequenceLength++]=result;
divider*=10;
if (divider==0) {
//We're solved.
break;
} }
} }
/*printf("%d:",divisor);
for (int i=0;i<sequenceLength;i++) {
printf("%d",sequence[i]);
}
printf("\n");*/
//We need to look at all combinations of possible repeating sequences.
//Starting from iteration loop of length 1, and an offset incrementing by 1, see if anything remains the same.
//Then do this for iteration loop of length 2, offset incrementing by 1, etc.
//Find it repeating at least 20 times to assume it repeats infinitely.
boolean isP = isPrime(primeList,primeListSize,divisor);
int*factors = getFactors(divisor-1);
int counter=0;
/*if (isP) {
printf("Possible length factors of (%d-1):",divisor);
while (factors[counter]!=0) {
printf("%d,",factors[counter]);
counter++;
}
printf("\n");
}*/
boolean matched=false;
if (sequenceLength==TARGET_REPEATS_REQUIRED) {
for (int checkLength=0;checkLength<TARGET_REPEATS_REQUIRED/10;checkLength++) {
if (isP&&!isFactor(factors,checkLength+1)) {continue;}
for (int offset=0;offset<sequenceLength+1-checkLength;offset++) {
int sequenceStore[checkLength+1];
int requiredCheckAmt=TARGET_REPEATS_REQUIRED/10-2;
for (int i=0;i<checkLength+1;i++) {
sequenceStore[i]=sequence[i+offset];
/*printf("%d:",divisor);
for (int i=0;i<checkLength+1;i++) {
printf("%d",sequenceStore[i]);
}
0010030090270812437311935807422266800401203610832497492477432296890672016048144433299899699097291875626880641925777331995987963891675025075225677031093279839518555667
0010030090270812437311935807422266800401203610832497492477432296890672016048144433299899699097291875626880641925777331995987963891675025075225677031093279839518555667
printf("\n");*/
}
//Now if this repeating sequence can be found requiredCheckAmt times, we're golden.
boolean allMatching=true;
int currentOffset=0;
int found=0;
while (requiredCheckAmt>0&&found<20) {
for (int i=0;i<checkLength+1;i++) {
//printf(" Compare %d to %d\n",sequence[offset+currentOffset+checkLength+1+i],sequenceStore[i]);
if (sequence[offset+currentOffset+checkLength+1+i]!=sequenceStore[i]) {
allMatching=false;
goto check;
}
}
requiredCheckAmt--;
currentOffset+=checkLength+1;
found++;
//printf("Found a repeat.");
}
check:
if (allMatching) {
printf("Longest repeating sequence for %d is of length %d: ",divisor,checkLength+1);
for (int i=0;i<checkLength+1;i++) {
printf("%d",sequenceStore[i]);
}
printf("\n");
if (longestCycleLength<checkLength+1) {
longestCycleLength=checkLength+1;
longestCycleDivisor=divisor;
for (int i=0;i<longestCycleLength;i++) {
longestSequence[i]=sequenceStore[i];
}
}
matched=true;
goto next;
}
}
}
} else {
matched=true;
printf("Longest repeating sequence for %d is of length 0.\n",divisor);
}
next:
free(factors);
if (!matched) {
printf("Not enough data to calculate longest repeating sequence for %d!\nQuitting...\n",divisor);
return 1;
}
divisor++;
} }
printf("\n\nThe largest repeating sequence length is from %d with %d in length.",longestCycleDivisor,longestCycleLength);
printf("\n\tSequence: ");
for (int i=0;i<longestCycleLength;i++) {
printf("%d",longestSequence[i]);
}
printf("\n");
return 0; return 0;
} }