parent
eb9733c9d7
commit
f0d9bbfaeb
@ -1,211 +1,40 @@ |
||||
#include <stdio.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. |
||||
|
||||
boolean isPrime(int*primeList,int primeListSize,int numb) { |
||||
for (int i=0;i<primeListSize;i++) { |
||||
if (numb==primeList[i]) { |
||||
return true; |
||||
} |
||||
} |
||||
return false; |
||||
} |
||||
Considering quadratics of the form: |
||||
|
||||
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; |
||||
} |
||||
, where and
|
||||
|
||||
boolean isFactor(int*factorList,int numb) { |
||||
int counter=0; |
||||
while (factorList[counter]!=0) { |
||||
if (numb==factorList[counter]) { |
||||
return true; |
||||
} |
||||
counter++; |
||||
} |
||||
return false; |
||||
} |
||||
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 . |
||||
|
||||
const int TARGET_REPEATS_REQUIRED=100000; |
||||
https://projecteuler.net/problem=27
|
||||
*/ |
||||
|
||||
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"); |
||||
while (fgetc(f)!='{'); |
||||
while (true) { |
||||
char c; |
||||
int digit=0; |
||||
while ((c=fgetc(f))!=',') { |
||||
digit*=10; |
||||
digit+=c-'0'; |
||||
}
|
||||
if (digit>=1000) { |
||||
break; |
||||
} else { |
||||
primeList=realloc(primeList,sizeof(int)*++primeListSize); |
||||
primeList[primeListSize-1]=digit; |
||||
} |
||||
int startMarker=0; |
||||
while (fgetc(f)!='{') { |
||||
startMarker++; |
||||
} |
||||
fseek(f,startMarker,SEEK_SET); |
||||
int n=0; |
||||
while (true) { |
||||
for (int a=1;a<1000;a++) { |
||||
for (int b=1;b<=1000;b++) { |
||||
|
||||
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; |
||||
} |
||||
Loading…
Reference in new issue