//函数fun功能:求n(n<10000)以内的所有四叶玫瑰数并逐个存放到result所指数组中,个数作为返回值.如果一个4位整数等于其各个位数字的4次方之和,则称该数为函数返回值. #include<stdio.h> #pragma warning (disable:4996) int fun(int n, int result[]) { ,j=; int a, b, c, d; ; i < n; i++) { a = i / ; b = (i % ) / ; c = (i %
本题来自 Project Euler 第21题:https://projecteuler.net/problem=21 ''' Project Euler: Problem 21: Amicable numbers Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a ≠ b
偶数除了二都不是素数 一个数 n 如果是合数,那么它的所有的因子不超过sqrt(n)--n的开方 int i, j, n = 10000; for (i = 3; i <= n; i += 2) { for (j = 2; j < (Math.Sqrt(i) + 1); j++) { if (i % j == 0) break; } if (j > Math.Sqrt(i)) Console.WriteLine("{0}", i); }
"Problem: To print in ascending order all primes less than 10000. Use an array of processes, SIEVE, in which each process inputs a prime from its predecessor and prints it. The process then inputs an ascending stream of numbers from its predecessor a