Project Euler practice

puzzles/q001To005.c

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+#include<stdio.h>
+#include<stdlib.h>
+#include<math.h>
+
+// for question3,4
+#define MAX(a,b) (a>b?a:b)
+
+
+int MaxInt(int a, int b)
+{
+	if(a>=b)
+		return a;
+	else
+		return b;
+}
+
+// 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.
+// SOLUTION: Brutal force might be a straight forward solution:)
+int question1()
+{
+	int sum = 0;
+	int i;
+	for( i=1; i<1000; i++)
+	{
+		if(i%3==0 || i%5==0)
+			sum +=i;
+	}
+	return sum;
+}
+
+// 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.
+// SOLUTION: The fibonacci sequence should be: odd, even, odd, odd, even, odd, odd, even...
+// So, the sum should contains elements with index 2,5,8,11......
+int question2()
+{
+	int sum=2;
+	int fab[3]={1,2,3};	// storing the fab sequence and only add the middle to "sum"
+	int index;
+	int count = 3;
+	while(1)
+	{
+		index=count%3;
+		fab[index]=fab[(count+1)%3]+fab[(count+2)%3];
+		if(fab[index]>4000000)
+			break;
+		count++;
+		if(index==1)
+			sum +=fab[index];
+	}
+	return sum;
+}
+
+
+// The prime factors of 13195 are 5, 7, 13 and 29.
+// What is the largest prime factor of the number 600851475143 ?
+// SOLUTION: There are quite a few algorithms to solve in in wiki: http://en.wikipedia.org/wiki/Integer_factorization
+// However, most of them are targetting at HUGE numbers with "100 digits".
+// So, considering the algorithm complexity and the input number in this puzzle,
+// it might be a good trade-off to considering 3 to n^.5 without even number.
+// The input for this question is: question3(3,600851475143);
+// The "begin" is just a small optimization cutting running time by 67%.
+long question3(long begin, long input)
+{
+	//int input = 600851475143;
+	long square;
+	long result = 0;
+	long i;
+
+	if(input==1)
+		return 1;
+	while(input%2==0)
+		input = input/2;
+	square = ceil(sqrt(input));
+	for(i=begin;i<square;i+=2)
+	{
+		if(input%i==0){
+			result = MAX(i,question3(i,ceil(input/i)));
+			break;
+		}
+	}
+	if(result==0)
+		result = MAX(2,input);
+	return result;
+}
+
+
+// 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.
+// SOLUTION: The most straight forward (but way too brutal) is to iteration two numbers from 999 to 100 and test the result.
+// To improve it, we can have to input variables to limit the range of searching
+// The input format for this Q should be "question4(999,100)"
+// Note that I have changed the MAX macro to MaxInt function for the macro performs wierd when I recursively call the function...
+int question4(int firstBegin, int lastEnd)
+{
+	int temp = firstBegin;
+	int product = 0;
+	int test1, test2, test3;
+	if(firstBegin<=lastEnd)
+	{
+		return 0;	// branch cut.
+	}
+	while(1)
+	{
+		product = firstBegin*temp;
+		if(product<100000){
+			product=0;
+			break;
+		}
+		test1 = (int)product/100000;
+		if(test1==product%10)
+		{
+			test2=((int)product/10000)%10;
+			if(test2==((int)product/10)%10)
+			{
+				test3=((int)product/1000)%10;
+				if(test3==((int)product/100)%10)
+					break;
+			}
+		}
+		if(temp<=lastEnd)
+		{
+			product=0;
+			break;	// cannot found
+		}
+		temp--;
+	}
+	product = MaxInt(product,(question4(firstBegin-1,temp)));
+	return product;
+}
+// IMPROVE QUESTION4
+int question4Improved()
+{
+	int result=0;
+	int a = 999, b, db;
+	int product, test1, test2, test3;
+	while(a>=100)
+	{
+		if(a%11==0)
+		{
+			b=999;
+			db=1;
+		} else {
+			b=990;
+			db=11;
+		}
+		while(b>=a)
+		{
+			product=a*b;
+			if(product<=result)
+				break;
+			// test if a*b is palindrome
+			test1 = (int)product/100000;
+			if(test1==product%10)
+			{
+				test2=((int)product/10000)%10;
+				if(test2==((int)product/10)%10)
+				{
+					test3=((int)product/1000)%10;
+					if(test3==((int)product/100)%10)
+					{
+						result=product;
+						break;
+					}
+				}
+			}
+			b=b-db;
+		}
+		a--;
+	}
+	return result;
+}
+
+
+// 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?
+// SOLUTION: the array can be divided as 2-n^.5 and n^.5-n. For the former part, we need to know how many powers for each prime, and for the latter part, we just need to multiply all of the primes.
+int question5()
+{
+	int range = 20;
+	int square = ceil(sqrt(range));
+	int n, power;
+	int result = 1;
+	int i, flag=0;
+
+	// find the power from 2 to square
+	for(n=2;n<square;n++)
+	{
+		// check if n is a prime
+		for(i=2;i<=floor(sqrt(n));i++)
+		{
+			if(n%i==0)
+			{
+				flag=1;
+				break;
+			}
+		}
+		if(flag)
+		{
+			flag=0;
+			continue;
+		}
+		// find how many powers are needed
+		power=2;
+		while(1)
+		{
+			if(pow(n,power)>=range)
+				break;
+			power++;
+		}
+		result = result*pow(n,(power-1));
+	}
+
+	// find all of the primes within the range
+	for(n=square;n<range;n++)
+	{
+		// check if n is a prime
+		for(i=2;i<=floor(sqrt(n));i++)
+		{
+			if(n%i==0)
+			{
+				flag=1;
+				break;
+			}
+		}
+		if(flag)
+		{
+			flag=0;
+			continue;
+		}
+		result=result*n;
+	}
+
+	return result;
+}
+
+
+
+int main(){
+	printf("result = %d.\n", question5());
+	return 0;
+}
+
+

puzzles/q006To010.c

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+#include<stdio.h>
+#include<stdlib.h>
+#include<math.h>
+
+// 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.
+// SOLUTION: This can be solved with the formula that calculate the sum of consecutive natural sequence: 1^2+2^2+...+n^2=n(n+1)(2n+1)/6, and the sum of consecutive natural numbers: 1+2+...+n=(n+1)n/2
+int question6()
+{
+	int n = 100;
+	int sum1, sum2;
+
+	sum1 = pow(n*(n+1)/2, 2);
+	sum2 = n*(n+1)*(2*n+1)/6;
+	return sum1-sum2;
+}
+
+// By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
+// What is the 10001st prime number?
+int question7()
+{
+	int count = 6;
+	int n=17;
+	int i;
+	int flag=0;
+
+	while(count!=10001)
+	{
+		for(i=2;i<=floor(sqrt(n));i++)
+		{
+			if(n%i==0)
+			{
+				flag=1;
+				break;
+			}
+		}
+		if(flag)
+			flag=0;
+		else{
+			count++;
+			printf("count=%d\tn=%d\n",count,n);
+		}
+		n+=2;
+	}
+	return n-2;	// the last round add 2 more to the final result
+}
+
+
+int main()
+{
+	printf("result = %d\n", question7());
+	return 0;
+}
+
+