Francisco Javier Coutiño / Project Euler cpp

01 - Multiplos 3 y 5/p01

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01 - Multiplos 3 y 5/p01.cpp

File was created	1	/*
	2
	3	Multiples of 3 and 5
	4	Problem 1
	5	If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
	6	Find the sum of all the multiples of 3 or 5 below 1000.
	7
	8	By: dbk
	9
	10	*/
	11
	12	#include <iostream>
	13	#include <stdio.h>
	14
	15	using namespace std;
	16
	17	int main()
	18	{
	19	int res = 0;
	20
	21	for (int i = 1; i < 1000; i++)
	22	{
	23	res += (i % 3 == 0 \|\| i % 5 == 0) ? i : 0;
	24	}
	25
	26	printf("Resultado: %i\n", res);
	27
	28	return 0;
	29	}

02 - Serie Fibonacci/p02

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02 - Serie Fibonacci/p02.cpp

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File was created	1	/*
	2	Even Fibonacci numbers
	3	Problem 2
	4	Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
	5
	6	1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
	7
	8	By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
	9
	10	by:dbk
	11
	12	*/
	13
	14	#include <stdio.h>
	15
	16	#define LIMITE 4000000
	17
	18	int main()
	19	{
	20
	21	int i;
	22	unsigned int a = 1;
	23	unsigned int b = 1;
	24	unsigned int res = 0;
	25	unsigned int tmp = 0;
	26
	27	do
	28	{
	29
	30	tmp = a;
	31	a = b;
	32	b += tmp;
	33
	34	if (b % 2 == 0) res += b;
	35
	36	}while(b < LIMITE);
	37
	38	printf("\nResultado final: %ld\n", res);
	39
	40	return 0;
	41	}

03 - Factor primo/p03

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03 - Factor primo/p03.cpp

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File was created	1	/*
	2
	3	Largest prime factor
	4	Problem 3
	5	Published on Friday, 2nd November 2001, 12:00 pm; Solved by 256800
	6	The prime factors of 13195 are 5, 7, 13 and 29.
	7
	8	What is the largest prime factor of the number 600851475143 ?
	9
	10	By: dbk
	11
	12	*/
	13	#include <stdio.h>
	14
	15	#define CALCULAR 600851475143
	16	#define MAX 1000000
	17
	18	int main(int argc, char const *argv[])
	19	{
	20
	21	long res = 0;
	22	long valor = CALCULAR;
	23
	24	for (int i = 2; i < MAX; i++)
	25	{
	26	while ( valor % i == 0)
	27	{
	28	valor /= i;
	29	res = i;
	30	}
	31
	32	}
	33
	34	printf("Mayor factor primo: %ld\n", res);
	35
	36	return 0;
	37	}

04 - Producto palindromo/itoaEjemplo.cpp

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File was created	1	#include <stdio.h>
	2	#include <stdlib.h>
	3
	4	int main ()
	5	{
	6	int i;
	7	char buffer [33];
	8	printf ("Enter a number: ");
	9	scanf ("%d",&i);
	10	itoa (i,buffer,10);
	11	printf ("decimal: %s\n",buffer);
	12	itoa (i,buffer,16);
	13	printf ("hexadecimal: %s\n",buffer);
	14	itoa (i,buffer,2);
	15	printf ("binary: %s\n",buffer);
	16	return 0;
	17	}

04 - Producto palindromo/p04

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04 - Producto palindromo/p04.cpp

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File was created	1	/*
	2	Largest palindrome product
	3	Problem 4
	4	A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
	5
	6	Find the largest palindrome made from the product of two 3-digit numbers.
	7
	8	By:dbk
	9	*/
	10	#include <stdio.h>
	11	#include <string.h>
	12
	13	using namespace std;
	14
	15	#define MAX 10000
	16
	17	bool validPalindromic(int);
	18	bool validPalindromicApuntador(int);
	19
	20	int main(int argc, char const *argv[])
	21	{
	22
	23	int a = 1;
	24	int b = 1;
	25
	26	int c = 0;
	27
	28	int res = 0;
	29
	30	do
	31	{
	32	a = 1;
	33	do
	34	{
	35	c = a * b;
	36	res = (validPalindromicApuntador(c) && c > res) ? c : res;
	37
	38	}while(a++ < MAX);
	39	}while(b++ < MAX);
	40	printf("Resultado final: %d\n", res);
	41	return 0;
	42	}
	43
	44
	45	bool validPalindromic(int valueCheck)
	46	{
	47	bool flagValid = true;
	48	char buffer[30];
	49	int counterDsc = 0;
	50	int counterAsc = 0;
	51	int sizeStringNumber = 0;
	52
	53	//Convierte el entero a una cadena
	54	sprintf(buffer, "%d", valueCheck);
	55	counterDsc = sizeStringNumber = strlen(buffer);
	56	while(sizeStringNumber > counterAsc) {
	57	flagValid &= (buffer[counterAsc++] == buffer[--counterDsc]) ? true : false;
	58	}
	59
	60	return flagValid;
	61	}
	62
	63	bool validPalindromicApuntador(int valueCheck)
	64	{
	65	char *ptrAsc;
	66	char *ptrDsc;
	67	char buffer[30];
	68	bool flagValid = true;
	69
	70	//Convierte el entero a una cadena
	71	sprintf(buffer, "%d", valueCheck);
	72	ptrAsc = buffer;
	73	ptrDsc = &buffer[strlen(buffer)];
	74
	75	while(ptrDsc >= ptrAsc) {
	76	flagValid &= ((ptrAsc++) == (--ptrDsc));
	77	}
	78	return flagValid;
	79	}
	80
	81	/*
	82	TEST DIR MEM Apuntadores
	83	printf("mem1:\t%p\n", ptrAsc);
	84	printf("mem2:\t%p\n", ptrDsc);
	85	*/

05 - Multiplo menor/p05

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05 - Multiplo menor/p05.cpp

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File was created	1	/*
	2	Smallest multiple
	3
	4	Problem 5
	5	2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
	6
	7	What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
	8	*/
	9	#include <stdio.h>
	10
	11	using namespace std;
	12
	13	#define MAX_VALUE 20
	14
	15
	16	int main(int argc, char const *argv[])
	17	{
	18	int i = 0;
	19	int incMultiplos = 0;
	20
	21	bool flgValid = true;
	22
	23	do
	24	{
	25	i++;
	26	flgValid = true;
	27	incMultiplos = MAX_VALUE;
	28
	29	while(incMultiplos > 0)
	30	{
	31	flgValid &= (i % incMultiplos-- == 0);
	32	if (!flgValid) break;
	33	}
	34
	35	}while(!flgValid);
	36
	37	printf("i: %d \t%x\n", i, flgValid);
	38
	39	return 0;
	40	}

06 - Suma de diferencia de cuadrados/p06

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06 - Suma de diferencia de cuadrados/p06.cpp

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File was created	1	/*
	2
	3	Sum square difference
	4	Problem 6
	5	The sum of the squares of the first ten natural numbers is,
	6
	7	12 + 22 + ... + 102 = 385
	8	The square of the sum of the first ten natural numbers is,
	9
	10	(1 + 2 + ... + 10)2 = 552 = 3025
	11	Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
	12
	13	Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
	14
	15	*/
	16
	17	#include <stdio.h>
	18	#include <math.h>
	19
	20	using namespace std;
	21
	22	#define LIMITE_RANGO 100
	23
	24	int main(int argc, char const *argv[])
	25	{
	26
	27	int i = 0;
	28	int res = 0;
	29	int valSum = 0;
	30	int valPow = 0;
	31
	32	while(i++ < LIMITE_RANGO)
	33	{
	34	valSum += i;
	35	valPow += pow(i, 2);
	36	}
	37
	38	res = pow(valSum, 2) - valPow;
	39	printf("Resultado final: %d\n", res);
	40
	41	return 0;
	42	}

p01.cpp

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File was created	1	#include <iostream>
	2
	3	using namespace std;
	4
	5	int main()
	6	{
	7	res = 0;
	8
	9	for (int i = 10; i > 0; i--)
	10	{
	11
	12	if (i % 3 == 0 \|\| i % 5)
	13	{
	14	res += i;
	15	}
	16
	17	}
	18
	19	cout << res << endl;
	20
	21	}

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Francisco Javier Coutiño / Project Euler cpp

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