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#25 completed

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Co-authored-by: sigonasr2 <sigonasr2@gmail.com>
main
sigonasr2, Sig, Sigo 2 years ago
parent 693ee1e4ae
commit 6ca700a8ef
7 changed files with 232 additions and 72 deletions
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  1. 7
      archives/24/src/main.c
  2. BIN
      archives/25/current
  3. 44
      archives/25/src/main.c
  4. 140
      archives/25/src/utils.c
  5. 14
      archives/25/src/utils.h
  6. BIN
      current
  7. 99
      src/main.c

7
archives/24/src/main.c
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@ -51,9 +51,6 @@ int factorial(int numb) {
int main(int argc,char**argv) {
//For permutations, we know that there are a total of N! permutations.
// Which means each individual set starts at N!/N. So we can skip ahead some permutations.
// For 10 numbers, we get 362,880 values per permutation set. So 1000000/362880 = 2. So we'll start at permutation two.
int setPermutationCount=factorial(10)/10;
int permutationCount=(1000000/setPermutationCount)*setPermutationCount;
int digits[10]={2,0,1,3,4,5,6,7,8,9};
@ -61,11 +58,11 @@ int main(int argc,char**argv) {
incrementDigit(digits);
if (allUniqueDigits(digits)) {
permutationCount++;
printf("%d:",permutationCount);
/*printf("%d:",permutationCount);
for (int i=0;i<10;i++) {
printf(" %d ",digits[i]);
}
printf("\n");
printf("\n");*/
}
}

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archives/25/src/main.c
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#include <stdio.h>
#include "utils.h"
#include <stdlib.h>
/*
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
https://projecteuler.net/problem=24
*/
int main(int argc,char**argv) {
struct String numb1 = BigNumber(1);
struct String numb2 = BigNumber(1);
int term=3;
while (numb2.length<1000) {
struct String oldString = numb1;
numb1 = numb2;
numb2 = add(oldString,numb1);
//free(oldString.str);
//printf("%d: %s\n",term,numb2.str);
term++;
}
printf("\n\nTerm %d has %d digits!",term,numb2.length);
return 0;
}

140
archives/25/src/utils.c
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#include "utils.h"
#include <stdio.h>
#include <stdlib.h>
struct String mult(struct String numb1, struct String numb2) {
struct String n1 = numb1;
struct String n2 = numb2;
byte carryover = 0;
if (numb2.length>numb1.length) {
n1=numb2;
n2=numb1;
}
int addends[n2.length][n1.length+1];
for (int i=0;i<n2.length;i++) {
for (int j=0;j<n1.length+1;j++) {
addends[i][j]=0;
}
}
for (int i=n2.length-1;i>=0;i--) {
carryover=0;
for (int j=n1.length-1;j>=0;j--) {
int mult = (n1.str[j]-'0')*(n2.str[i]-'0')+((carryover!=0)?carryover:0);
//printf(" %d/%d\n",mult,carryover);
carryover=0;
if (mult>=10) {
carryover=mult/10;
mult=mult%10;
}
addends[(n2.length-1)-i][j+1]=mult;
}
if (carryover>0) {
addends[(n2.length-1)-i][0]=carryover;
}
}
//printIntDoubleArr(n2.length,n1.length+1,addends);
struct String sum = {1,"0"};
for (int i=0;i<n2.length;i++) {
char val[n1.length+1+i];
for (int j=0;j<n1.length+1+i;j++) {
val[j]='0';
}
for (int j=0;j<n1.length+1;j++) {
val[j]=addends[i][j]+'0';
}
sum=add((struct String){n1.length+1+i,val},sum);
//printf("%s\n",sum.str);
}
if (sum.str[0]=='0') {
char*newStr=malloc(sum.length-1);
for (int i=1;i<sum.length;i++) {
newStr[i-1]=sum.str[i];
}
free(sum.str);
sum=(struct String){sum.length-1,newStr};
}
//printf("%s",sum.str);
return sum;
}
struct String add(struct String numb1, struct String numb2){
//printf("%s %s\n",numb1.str,numb2.str);
byte carryover=0;
int digitCount=0;
char*str = malloc(1);
//str[0]='\0';
if (numb1.length>=numb2.length) {
for (int offset=0;offset<numb1.length;offset++) {
str = realloc(str,++digitCount);
//printf("Digit count is now %d\n",digitCount);
if (numb2.length>offset) {
//printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]);
int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0);
if (sum>=10) {
carryover=1;
sum-=10;
}
str[offset]=sum+'0';
} else {
str[offset]=numb1.str[numb1.length-offset-1]+((carryover>0)?carryover--:0);
}
//str[offset+1]='\0';
}
} else {
for (int offset=0;offset<numb2.length;offset++) {
str = realloc(str,++digitCount);
//printf("Digit count is now %d\n",digitCount);
if (numb1.length>offset) {
//printf("%c %c\n",numb1.str[numb1.length-offset-1],numb2.str[numb2.length-offset-1]);
int sum=((numb1.str[numb1.length-offset-1]-'0')+(numb2.str[numb2.length-offset-1]-'0'))+((carryover>0)?carryover--:0);
if (sum>=10) {
carryover=1;
sum-=10;
}
str[offset]=sum+'0';
} else {
str[offset]=numb2.str[numb2.length-offset-1]+((carryover>0)?carryover--:0);
}
//str[offset+1]='\0';
}
}
if (carryover>0) {
str = realloc(str,++digitCount);
str[digitCount-1]='1';
//str[digitCount]='\0';
}
for (int i=0;i<digitCount/2;i++) {
char c = str[i];
char c2 = str[digitCount-i-1];
str[digitCount-i-1]=c;
str[i]=c2;
}
str = realloc(str,digitCount+1);
str[digitCount]='\0';
struct String newStr = {digitCount,str};
return newStr;
}
void printLongDoubleArr(int a,int b,long doubleArr[a][b]) {
for (int i=0;i<a;i++) {
for (int j=0;j<b;j++) {
printf("%ld\t",doubleArr[i][j]);
}
printf("\n");
}
}
void printIntDoubleArr(int a,int b,int doubleArr[a][b]) {
for (int i=0;i<a;i++) {
for (int j=0;j<b;j++) {
printf("%d\t",doubleArr[i][j]);
}
printf("\n");
}
}
struct String createBigNumber(char*numb) {
int marker=0;
while (numb[marker++]!='\0');
return (struct String){marker-1,numb};
}

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archives/25/src/utils.h
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#define true 1
#define false 0
#define boolean char
#define byte char
struct String{
int length;
char*str;
};
struct String add(struct String numb1, struct String numb2);
struct String mult(struct String numb1, struct String numb2);
void printLongDoubleArr(int a,int b,long doubleArr[a][b]);
void printIntDoubleArr(int a,int b,int doubleArr[a][b]);
struct String createBigNumber(char*numb);
#define BigNumber(X) createBigNumber(#X)

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src/main.c
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@ -1,79 +1,44 @@
#include <stdio.h>
#include "utils.h"
#include <stdlib.h>
/*
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
https://projecteuler.net/problem=24
*/
void carryover(int*arr,int digit) {
arr[digit]=0;
if (arr[digit-1]==9) {
carryover(arr,digit-1);
} else {
arr[digit-1]++;
}
}
void incrementDigit(int*arr) {
int val = arr[9];
if (arr[9]==9) {
carryover(arr,9);
} else {
arr[9]++;
}
}
boolean allUniqueDigits(int*arr) {
boolean digits[10] = {};
for (int i=0;i<10;i++) {
if (!digits[arr[i]]) {
digits[arr[i]]=true;
} else {
return false;
}
}
return true;
}
int factorial(int numb) {
int final=1;
for (int i=numb;i>=1;i--) {
final*=i;
}
return final;
}
int main(int argc,char**argv) {
//For permutations, we know that there are a total of N! permutations.
// Which means each individual set starts at N!/N. So we can skip ahead some permutations.
// For 10 numbers, we get 362,880 values per permutation set. So 1000000/362880 = 2. So we'll start at permutation two.
int setPermutationCount=factorial(10)/10;
int permutationCount=(1000000/setPermutationCount)*setPermutationCount;
int digits[10]={2,0,1,3,4,5,6,7,8,9};
while ((digits[0]!=9||digits[1]!=8||digits[2]!=7||digits[3]!=6||digits[4]!=5||digits[5]!=4||digits[6]!=3||digits[7]!=2||digits[8]!=1||digits[9]!=0)&&permutationCount!=999999) {
incrementDigit(digits);
if (allUniqueDigits(digits)) {
permutationCount++;
printf("%d:",permutationCount);
for (int i=0;i<10;i++) {
printf(" %d ",digits[i]);
}
printf("\n");
}
struct String numb1 = BigNumber(1);
struct String numb2 = BigNumber(1);
int term=3;
while (numb2.length<1000) {
struct String oldString = numb1;
numb1 = numb2;
numb2 = add(oldString,numb1);
//free(oldString.str);
//printf("%d: %s\n",term,numb2.str);
term++;
}
printf("\n\nThe one millionth lexicographic permutation is: \n\t");
for (int i=0;i<10;i++) {
printf("%d",digits[i]);
}
printf("\n\nTerm %d has %d digits!",term,numb2.length);
return 0;
}
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