题意:在20×20方阵中从起点出发只允许向右或向下移动到达终点的路径有多少条. 思路:每次只能向右或者向下,总共 40 步,也就是 40 步中每一步都有两种选择,也就是 C (40 , 20) . 为什么能在计算分子的时候不断约分掉分母?首先,组合数是整数,也就是说到最后分子一定能整除分母.我们使用 m 记录当前还没有被约分掉的最大的数,如果分子能够整除掉 m 就进行约分并且 m 更新为下一个等待约分的值.这样做就可以避免在计算组合数中导致的数据溢出问题! /******************…

这是一道很综合的计数问题,对于思维的全面性,解法的过渡性,代码能力,细节处理,计数问题中的各种算法,像gcd.容斥.类欧几里德算法都有考察.在省选模拟赛中做到了这题,然而数据范围是n,m小于等于1000.首先有一个O(n^4m^4)的暴力.然后开始计数,思路是:答案等于任取4个点的方案数+2*取4个点不为凸的方案.前一部分相对来说容易统计,先用组合数算所有的,再把存在3点.4点共线的矩形的贡献减掉就好了.这里用到了矩形框的思路,利用了容斥,而且在计数的时候用gcd作为工具,这个思路下面还会用到,…

组合数,2n中选n个.向右走有n步,向下走有n步.共2n步.有n步是向右走的,计算向右走的这n步的所有情况,即C(2n,n). 或者,每一步,只能从右边或者上边走过来,只有这两种情况,即step[i][j] = step[i-1][j]+step[i][j-1],递推即可. #include <iostream> using namespace std; typedef long long ll; ll C(ll n, ll m) { ll res = 1; for(ll i = 1; i &…

本题来自 Project Euler 第15题:https://projecteuler.net/problem=15 ''' Project Euler: Problem 15: Lattice paths Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right…

Starting in the top left corner of a 22 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner. How many such routes are there through a 2020 grid? 题目大意: 从一个22网格的左上角开始,有6条(不允许往回走)通往右下角的路. 对于2020…

In Problem 42 we dealt with triangular problems, in Problem 44 of Project Euler we deal with pentagonal number, I can only wonder if we have to deal with septagonal numbers in Problem 46. Anyway the problem reads Pentagonal numbers are generated by t…

Project Euler: 欧拉计划是一系列挑战数学或者计算机编程问题,解决这些问题需要的不仅仅是数学功底. 启动这一项目的目的在于,为乐于探索的人提供一个钻研其他领域并且学习新知识的平台,将这一平台打造一个有趣和休闲 的环境. 项目主页:https://projecteuler.net 第一题 Multiples of 3 and 5 If we list all the natural numbers below 10 that are multiples of 3 or 5, we ge…

本题来自 Project Euler 第22题:https://projecteuler.net/problem=22 ''' Project Euler: Problem 22: Names scores Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into al…

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12题:https://projecteuler.net/problem=12 # Project Euler: Problem 12: Highly divisible triangular number # The sequence of triangle numbers is generated by adding the natural numbers. # So the 7th triangle number would be 1 + 2 + 3…

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 ×…

上一次接触 project euler 还是2011年的事情,做了前三道题,后来被第四题卡住了,前面几题的代码也没有保留下来. 今天试着暴力破解了一下,代码如下: (我大概是第 172,719 个解出这道题的人) program 4 A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.…

开始做 Project Euler 的练习题.网站上总共有565题,真是个大题库啊! # Project Euler, Problem 1: Multiples of 3 and 5 # 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…

题意:三个正整数a + b + c = 1000,a*a + b*b = c*c.求a*b*c. 解法:可以暴力枚举,但是也有数学方法. 首先,a,b,c中肯定有至少一个为偶数,否则和不可能为以上两个等式均不会成立.然后,不可能a,b为奇c为偶,否则a*a%4=1, b*b%4=1, 有(a*a+b*b) %4 = 2,而c*c%4 = 0.也就是说,a和b中至少有一个偶数. 这是勾股数的一个性质,a,b中至少有一个偶数. 然后,解决过程见下(来自project euler的讨论): tag:m…

project euler 169 题目链接:https://projecteuler.net/problem=169 参考题解:http://tieba.baidu.com/p/2738022069 #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define mp make_pair #define pb push_back #define rep(i, a, b) for(i…

题目要求是: The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832. 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319…

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…

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19题:https://projecteuler.net/problem=19 ''' How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)? Answer: 171 ''' from datetime import * firstDay = date(1901,1,1) lastDay = date(…

本题来自 Project Euler 第17题:https://projecteuler.net/problem=17 ''' Project Euler 17: Number letter counts If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. If all th…

本题来自 Project Euler 第14题:https://projecteuler.net/problem=14 ''' Project Euler: Problem 14: Longest Collatz sequence The following iterative sequence is defined for the set of positive integers: n → n/2 (n is even) n → 3n + 1 (n is odd) Using the rule…

本题来自 Project Euler 第13题:https://projecteuler.net/problem=13 # Project Euler: Problem 13: Large sum # Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. # Answer: 5537376230 numbers = '''371072875339021027987979982…

本题来自 Project Euler 第10题:https://projecteuler.net/problem=10 # Project Euler: Problem 10: Summation of primes # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. # Answer: 142913828922 def f(x):…

本题来自 Project Euler 第9题:https://projecteuler.net/problem=9 # Project Euler: Problem 9: Special Pythagorean triplet # A Pythagorean triplet is a set of three natural numbers, # a < b < c, for which, a**2 + b**2 = c**2 # For example, 3**2 + 4**2 = 9 +…

本题来自 Project Euler 第8题:https://projecteuler.net/problem=8 # Project Euler: Problem 8: Largest product in a series # The four adjacent digits in the 1000-digit number # that have the greatest product are 9 × 9 × 8 × 9 = 5832. # Find the thirteen adjac…

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6题:https://projecteuler.net/problem=6 # Project Euler: Problem 6: Sum square difference # 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 natur…

本题来自 Project Euler 第5题:https://projecteuler.net/problem=5 # Project Euler: Problem 5: Smallest multiple # 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 num…

本题来自 Project Euler 第4题:https://projecteuler.net/problem=4 # Project Euler: Problem 4: Largest palindrome product # A palindromic number reads the same both ways. # The largest palindrome made from the product # of two 2-digit numbers is 9009 = 91 × 9…

本题来自 Project Euler 第3题:https://projecteuler.net/problem=3 # Project Euler: Problem 3: Largest prime factor # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number 600851475143 ? # Answer: 6857 a = 6008514751…