摘要： 就是大数相加，100个50位的数相加，取和的前10位Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.3710728753390210279879799822083759024651013574025046376937677490009712648124896970078050417018260538743249861995247410594742333095130581237266173096299194221336357416157252243056330181107 阅读全文

摘要： 求第一个有500个除数因子的三角数，具体题目见原文：The sequence of triangle numbers is generated by adding the natural numbers. So the 7thtriangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...Let us list the factors of the first seven triangle numbers:1 阅读全文

摘要： 对一个20*20的矩阵，求斜对角或者竖行或者横行连续4个数字的最大乘积思想：只会很笨的解法，读入文件到一个二维数组，然后对数组中的每个数字，进行4种可能的计算。。。官网没参考答案。。。代码：View Code 1 private static int gridproduct(String s) { 2 int product = 0; 3 int[][] a = readFile(s); 4 int rows = a.length; 5 int cols = a[0].length; 6 for (int i =... 阅读全文

摘要： 前面腾讯面试这题时，很多概念都搞混了，过来反省，好好研究了下KMP算法，把它实现。大体过程数据结构上都说得很明白，这里说说自己的感悟吧：主要是next数组的作用：现有一个主串s和模式串t，KMP算法会首先对模式串t求一个辅助数组next，有个这个next后我们每次匹配时主串的下标i就不用回溯，模式串的下标j回溯到next[j]的位置。看似next值越大越好，其实不是这样，next值当然是为0，为-1时最好。这个next数值k的意思表示当si!=tj时，那么tj-ktj-k+1...tj-1和主串的s0s1..sk-1是匹配的，那么其实需要回溯k位置，即主串回溯到i-k位置和t0比较。当然，由于 阅读全文

摘要： 求小于N的所有素数的和/** * The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. * * Find the sum of all the primes below two million. */思路：个人只会傻逼的遍历判断这个数是否为素数，然后迭加，代码如下：View Code private static long sumPrime(int N) { if (N < 2) return 0; if (N == 2) return 2; l... 阅读全文

摘要： 求和为1000毕达哥斯加三元组，其实就是余弦定理的三个数。。。A Pythagorean triplet is a set of three natural numbers, a b c, for * which, * * a2 + b2 = c2 For example, 32 + 42 = 9 + 16 = 25 = 52. * * There exists exactly one Pythagorean triplet for which a + b + c = 1000. * Find the product abc.思路：这道题木有什么好的思路，个人只会遍历，官网的方法... 阅读全文

摘要： 找出如下1000个整数中5个连续的数最大的乘积/** * Find the greatest product of five consecutive digits in the 1000-digit * number. * * 73167176531330624919225119674426574742355349194934 * 96983520312774506326239578318016984801869478851843 * 85861560789112949495459501737958331952853208805511 * 1254069874715852386... 阅读全文

摘要： 求第10001个素数/** * 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 10 001st prime number? */代码：private static int nthPrime(int N) { if (N == 1) return 2; int k = 1; while (--N > 0) { k = k + 2; while (!isPrime(k)) ... 阅读全文

摘要： 求n个数和的平和与n的平方和的差/** * 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 nu... 阅读全文

摘要： 求能够被1-20中任意一个数整除的最小正整数/** * 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? */代码：private static long smallMul(int N) { long r =... 阅读全文