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Projecteuler Problem 1-10

2008-08-25 14:16 309 查看
Problem1
原题：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。

分析：求出1000（不包括1000）以下3或5倍数的和。

可以直接循环求，也可以用抽屉原理及等差数列求和公式来。
代码：public static int problem1(int max) {
int sum3 = sumOfArray(3, (max - 1) / 3, 3);
int sum5 = sumOfArray(5, (max - 1) / 5, 5);
int sum15 = sumOfArray(15, (max - 1) / 15, 15);

return sum3 + sum5 - sum15;
}

/**
* Calculate the sum of a arithmetic progression.
*
* @param a1 : the first term
* @param n: the length of array
* @param d:common difference
*/
private static int sumOfArray(int a1, int n, int d) {
return n * a1 + n * (n - 1) * d / 2;
}
Problem2
原题：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, ...

Find the sum of all the even-valued terms in the sequence which do not exceed four million.
分析：求出Fibonacci数列中，小于4000000且为偶数的和

慢慢来呗！
代码：public static int problem2(int max) {
int pre = 1;
int next = 2;
int sum = 0;
while (next <= max) {
if (next % 2 == 0) {
sum += next;
}
System.out.println(pre + "/t" + next + "/t" + sum);
int temp = pre;
pre = next;
next += temp;
}
return sum;
}
Problem3
原题：The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?
分析：就是求600851475143最大质因数。

代码：public static long problem3(long num) {
while (true) {
long temp = 0;
for (long i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
temp = (num / i);
break;
}
}
if (temp == 0) {
break;
}

num = temp;
}
return num;
}
Problem4
原题：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

分析：palindromic number:回文数

题目求的是两个三位数积中最大的回文数，邪恶的暴力。

代码：public static int problem4() {
int result = -1;
for (int i = 999; i >= 100; i--) {
for (int j = i; j >= 100; j--) {
if (isPalindromic(i * j)) {
if (i * j > result) {
result = i * j;
}
}
}
}
return result;
}

private static boolean isPalindromic(int num) {
char[] cs = String.valueOf(num).toCharArray();
for (int i = 0; i < cs.length / 2; i++) {
if (cs[i] != cs[cs.length - i - 1]) {
return false;
}
}
return true;
}
Problem5
原题：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 number that is evenly divisible by all of the numbers from 1 to 20?

分析：其实题目求的就是1～20的最小公倍数。

由于数字比较小，且这个数必须是1～20中所有质数的公倍数，所以暴力吧！
代码：public static int problem5() {
int begin = 1 * 3 * 5 * 7 * 11 * 13 * 17 * 19;
for (int i = begin;; i += begin) {
boolean check = true;
for (int j = 1; j <= 20; j++) {
if (i % j != 0) {
check = false;
}
}
if (check) {
System.out.println(i / begin);
return i;
}
}
}
Problem6
原题：The sum of the squares of the first ten natural numbers is,

1^2 + 2^2 + ... + 10^2 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)^2 = 55^2 = 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.

分析：求的是：1到100的平方和减去1到100和的平方的差

代码：public static int problem6() {
int result0 = 0;
int result1 = 0;
for (int i = 1; i <= 100; i++) {
result0 += (i * i);
result1 += i;
}
return result0 - result1 * result1;
}
Problem7
原题：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?

分析：求第10001个质数
代码：public static int p7() {
int length = 10001;
List<Integer> array = new ArrayList<Integer>(length);
array.add(2);
int temp = 3;
while (array.size() < length) {
boolean isPrime = true;
for (Integer i : array) {
if (temp % i == 0) {
isPrime = false;
}
}
if (isPrime) {
array.add(temp);
}
temp++;
}
return array.get(length - 1);
}
Problem8
原题：Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450
分析：在上面1000个数字中，求出最大连续5个数的积
代码：public static int p8() {
String s = "73167176531330624919225119674426574742355349194934"
+ "96983520312774506326239578318016984801869478851843"
+ "85861560789112949495459501737958331952853208805511"
+ "12540698747158523863050715693290963295227443043557"
+ "66896648950445244523161731856403098711121722383113"
+ "62229893423380308135336276614282806444486645238749"
+ "30358907296290491560440772390713810515859307960866"
+ "70172427121883998797908792274921901699720888093776"
+ "65727333001053367881220235421809751254540594752243"
+ "52584907711670556013604839586446706324415722155397"
+ "53697817977846174064955149290862569321978468622482"
+ "83972241375657056057490261407972968652414535100474"
+ "82166370484403199890008895243450658541227588666881"
+ "16427171479924442928230863465674813919123162824586"
+ "17866458359124566529476545682848912883142607690042"
+ "24219022671055626321111109370544217506941658960408"
+ "07198403850962455444362981230987879927244284909188"
+ "84580156166097919133875499200524063689912560717606"
+ "05886116467109405077541002256983155200055935729725"
+ "71636269561882670428252483600823257530420752963450";
char[] cs = s.toCharArray();
int result = 1;
int index = 0;
for (int i = 0; i < s.length() - 5; i++) {
int num1 = cs[i] - '0';
int num2 = cs[i + 1] - '0';
int num3 = cs[i + 2] - '0';
int num4 = cs[i + 3] - '0';
int num5 = cs[i + 4] - '0';
if (num1 * num2 * num3 * num4 * num5 > result) {
result = num1 * num2 * num3 * num4 * num5;
index = i;
}
}
System.out.println(cs[index] + "" + cs[index + 1] + "" + cs[index + 2]
+ "" + cs[index + 3] + "" + cs[index + 4]);
return result;
}
Problem9
原题：A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a^2 + b^2 = c2

For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.

Find the product abc.
分析：求满足以下条件三个整数的积：

1,构成勾股数；2,和为1000.

无耻地暴力吧
代码：public static void p9() {
for (int a = 3; a <= 333; a++) {
for (int b = a + 1; b <= 666; b++) {
int c = 1000 - a - b;
if (c <= b) {
break;
}

if (a * a + b * b == c * c) {
System.out.printf("%d %d %d", a, b, c);
System.out.println();
System.out.printf("%d + %d = %d", a * a, b * b, c * c);
System.out.println();
System.out.printf("%d", a * b * c);
}
}
}
}
Problem10
原题：The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.
分析：求2百万以下质数的和

代码：public static long p10() {
List<Integer> array = new ArrayList<Integer>();
array.add(2);
int temp = 3;
int max = 2000000;
long sum = 2;
for (; temp < max; temp += 2) {
boolean isPrime = true;
int sqrt = (int) Math.sqrt(temp);
for (Integer i : array) {
if (temp % i == 0) {
isPrime = false;
}
if (i > sqrt) {
break;
}
}
if (isPrime) {
array.add(temp);
sum += temp;
}
}
return sum;
}
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