题意:判断一个数 N 的各个位数阶乘之和是否为其本身,找出所有符合要求的数然后求和 思路:此题思路跟 30 题相同,找到枚举上界 10 ^ n <= 9! × n ,符合要求的 n < 6 ,因此选择 9! × 6 作为上界 /************************************************************************* > File Name: euler034.c > Author: WArobot > Blog: ht…

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. #include <iostream> #include <v…

题意:49/98是一个有趣的分数,因为可能在化简时错误地认为,等式49/98 = 4/8之所以成立,是因为在分数线上下同时抹除了9的缘故.分子分母是两位数且分子小于分母的这种有趣的分数有4个,将这四个分数的乘积写成最简分数,求此时分母的值. 思路:直接枚举判断即可,需要注意 11/22 这种类型的数 /************************************************************************* > File Name: euler033.c…

题意:判断一个数 N 的每一位的5次方的和是否为其本身 ,求出所有满足条件的数的和 思路:首先设这个数 N 为 n 位,可以简单的判断一下这个问题的上界 10 ^ n <= 9 ^ 5 × n,可以解得满足条件的最小解为 5 ,所以取 9 ^ 5 × 6 = 354294 作为枚举的上界 /************************************************************************* > File Name: euler030.c >…

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. 题目大意: 145 是一个奇怪的数字, 因为 1! + 4! + 5!…

本题来自 Project Euler 第20题:https://projecteuler.net/problem=20 ''' Project Euler: Problem 20: Factorial digit sum n! means n × (n − 1) × ... × 3 × 2 × 1 For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! i…

本题来自 Project Euler 第16题:https://projecteuler.net/problem=16 ''' Project Euler 16: Power digit sum 2**15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 2**1000? Answer: 1366 ''' print(sum(int(i…

本题来自 Project Euler 第11题:https://projecteuler.net/problem=11 # Project Euler: Problem 10: Largest product in a grid # In the 20×20 grid below, four numbers along a diagonal line have been marked in red. # The product of these numbers is 26 × 63 × 78 ×…

本题来自 Project Euler 第7题:https://projecteuler.net/problem=7 # Project Euler: Problem 7: 10001st prime # 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? # Answer…

本题来自 Project Euler 第2题:https://projecteuler.net/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…