Agregando problemas resueltos

Francisco Javier Coutiño
1 parent 4faa017546
Showing 14 changed files with 312 additions and 0 deletions Show diff stats
01 - Multiplos 3 y 5/p01
01 - Multiplos 3 y 5/p01.cpp
02 - Serie Fibonacci/p02
02 - Serie Fibonacci/p02.cpp
03 - Factor primo/p03
03 - Factor primo/p03.cpp
04 - Producto palindromo/itoaEjemplo.cpp
04 - Producto palindromo/p04
04 - Producto palindromo/p04.cpp
05 - Multiplo menor/p05
05 - Multiplo menor/p05.cpp
06 - Suma de diferencia de cuadrados/p06
06 - Suma de diferencia de cuadrados/p06.cpp
p01.cpp
@@ -0,0 +1,29 @@
+/*
+
+Multiples of 3 and 5
+Problem 1
+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.
+Find the sum of all the multiples of 3 or 5 below 1000.
+
+By: dbk
+
+*/
+
+#include <iostream>
+#include <stdio.h>
+
+using namespace std;
+
+int main()
+{
+	int res = 0;
+
+	for (int i = 1; i < 1000; i++)
+	{
+		res += (i % 3 == 0 || i % 5 == 0) ? i : 0;
+	}
+
+	printf("Resultado: %i\n", res);
+
+	return 0;
+}
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@@ -0,0 +1,41 @@
+/*
+Even Fibonacci numbers
+Problem 2
+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:
+
+1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+
+by:dbk
+
+*/
+
+#include <stdio.h>
+
+#define LIMITE 4000000
+
+int main()
+{
+
+	int i;
+	unsigned int a = 1;
+	unsigned int b = 1;
+	unsigned int res = 0;
+	unsigned int tmp = 0;
+
+	do
+	{
+
+		tmp = a;
+		a = b;
+		b += tmp;
+
+		if (b % 2 == 0) res += b;
+
+	}while(b < LIMITE);
+
+	printf("\nResultado final: %ld\n", res);
+
+	return 0;
+}
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+/*
+
+Largest prime factor
+Problem 3
+Published on Friday, 2nd November 2001, 12:00 pm; Solved by 256800
+The prime factors of 13195 are 5, 7, 13 and 29.
+
+What is the largest prime factor of the number 600851475143 ?
+
+By: dbk
+
+*/
+#include <stdio.h>
+
+#define CALCULAR 600851475143
+#define MAX 1000000
+
+int main(int argc, char const *argv[])
+{
+
+	long res = 0;
+	long valor = CALCULAR;
+
+	for (int i = 2; i < MAX; i++)
+	{
+		while ( valor % i == 0)
+		{
+			valor /= i;
+			res = i;
+		}
+
+	}
+
+	printf("Mayor factor primo: %ld\n", res);
+
+	return 0;
+}
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@@ -0,0 +1,17 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+int main ()
+{
+  int i;
+  char buffer [33];
+  printf ("Enter a number: ");
+  scanf ("%d",&i);
+  itoa (i,buffer,10);
+  printf ("decimal: %s\n",buffer);
+  itoa (i,buffer,16);
+  printf ("hexadecimal: %s\n",buffer);
+  itoa (i,buffer,2);
+  printf ("binary: %s\n",buffer);
+  return 0;
+}
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@@ -0,0 +1,85 @@
+/*
+Largest palindrome product
+Problem 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.
+
+Find the largest palindrome made from the product of two 3-digit numbers.
+
+By:dbk
+*/
+#include <stdio.h>
+#include <string.h>
+
+using namespace std;
+
+#define MAX 10000
+
+bool validPalindromic(int);
+bool validPalindromicApuntador(int);
+
+int main(int argc, char const *argv[])
+{
+
+	int a = 1;
+	int b = 1;
+
+	int c = 0;
+
+	int res = 0;
+
+	do
+	{
+		a = 1;
+		do
+		{
+			c = a * b;
+			res = (validPalindromicApuntador(c) && c > res) ? c : res;
+
+		}while(a++ < MAX);
+	}while(b++ < MAX);
+	printf("Resultado final: %d\n", res);
+	return 0;
+}
+
+
+bool validPalindromic(int valueCheck)
+{
+	bool flagValid = true;
+	char buffer[30];
+	int counterDsc = 0;
+	int counterAsc = 0;
+	int sizeStringNumber = 0;
+
+	//Convierte el entero a una cadena
+	sprintf(buffer, "%d", valueCheck);
+	counterDsc = sizeStringNumber = strlen(buffer);
+	while(sizeStringNumber > counterAsc) {
+		flagValid &= (buffer[counterAsc++] == buffer[--counterDsc]) ? true : false;
+	}
+
+	return flagValid;
+}
+
+bool validPalindromicApuntador(int valueCheck)
+{
+	char *ptrAsc;
+	char *ptrDsc;
+	char buffer[30];
+	bool flagValid = true;
+
+	//Convierte el entero a una cadena
+	sprintf(buffer, "%d", valueCheck);
+	ptrAsc = buffer;
+	ptrDsc = &buffer[strlen(buffer)];
+
+	while(ptrDsc >= ptrAsc) {
+		flagValid &= (*(ptrAsc++) == *(--ptrDsc));
+	}
+	return flagValid;
+}
+
+/*
+	TEST DIR MEM Apuntadores
+	printf("mem1:\t%p\n", ptrAsc);
+	printf("mem2:\t%p\n", ptrDsc);
+*/
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@@ -0,0 +1,40 @@
+/*
+Smallest multiple
+
+Problem 5
+	2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
+
+	What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
+*/
+#include <stdio.h>
+
+using namespace std;
+
+#define MAX_VALUE 20
+
+
+int main(int argc, char const *argv[])
+{
+	int i = 0;
+	int incMultiplos = 0;
+
+	bool flgValid = true;
+
+	do
+	{
+		i++;
+		flgValid = true;
+		incMultiplos = MAX_VALUE;
+
+		while(incMultiplos > 0)
+		{
+		    flgValid &= (i % incMultiplos-- == 0);
+		    if (!flgValid) break;
+		}
+
+	}while(!flgValid);
+
+	printf("i: %d \t%x\n", i, flgValid);
+
+	return 0;
+}
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@@ -0,0 +1,42 @@
+/*
+
+Sum square difference
+Problem 6
+The sum of the squares of the first ten natural numbers is,
+
+12 + 22 + ... + 102 = 385
+The square of the sum of the first ten natural numbers is,
+
+(1 + 2 + ... + 10)2 = 552 = 3025
+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.
+
+Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
+
+*/
+
+#include <stdio.h>
+#include <math.h>
+
+using namespace std;
+
+#define LIMITE_RANGO 100
+
+int main(int argc, char const *argv[])
+{
+
+	int i = 0;
+	int res = 0;
+	int valSum = 0;
+	int valPow = 0;
+
+	while(i++ < LIMITE_RANGO)
+	{
+		valSum += i;
+		valPow += pow(i, 2);
+	}
+
+	res = pow(valSum, 2) - valPow;
+	printf("Resultado final: %d\n", res);
+
+	return 0;
+}
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@@ -0,0 +1,21 @@
+#include <iostream>
+
+using namespace std;
+
+int main()
+{
+	res = 0;
+
+	for (int i = 10; i > 0; i--)
+	{
+
+		if (i % 3 == 0 || i % 5)
+		{
+			res += i;
+		}
+
+	}
+
+	cout << res << endl;
+
+}
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...	...	@@ -0,0 +1,29 @@
	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	+}
0	30	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,41 @@
	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	+}
0	42	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,37 @@
	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	+}
0	38	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,17 @@
	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	+}
0	18	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,85 @@
	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	+*/
0	86	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,40 @@
	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	+}
0	41	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,42 @@
	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	+}
0	43	\ No newline at end of file
...	...
...	...	@@ -0,0 +1,21 @@
	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	+}
0	22	\ No newline at end of file
...	...