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euler
[Project
Euler
] Problem 13
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. 37107287533902102798797998220837590246510135740250 46376937677490009712648124896970078050417018260538 74324986
·
2015-11-01 13:31
project
[Project
Euler
] Problem 12
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28,
·
2015-11-01 13:30
project
[Project
Euler
] Problem 11
In the 2020 grid below, four numbers along a diagonal line have been marked in red. 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56
·
2015-11-01 13:29
project
[Project
Euler
] Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. 求所以小于2,000,000的素数之和 对于素数,我还是利用第七问中的查表法 我们需要保留在表中的素数是所有小于sqrt(2000000)的素数 也就是小于1415的素数
·
2015-11-01 13:29
project
[Project
Euler
] Problem 9
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 +
·
2015-11-01 13:28
project
[Project
Euler
] Problem 7
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? 求第10001个素数。 求素数有很多种方法 最直接的方法就是看这个数能不能整除2~sqrt(该数); 而查表法能做更少的
·
2015-11-01 13:27
project
[Project
Euler
] Problem 8
Find the greatest product of five consecutive digits in the 1000-digit number. 73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949
·
2015-11-01 13:27
project
[Project
Euler
] Problem 6
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
·
2015-11-01 13:26
project
[Project
Euler
] Problem 5
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 t
·
2015-11-01 13:25
project
[Project
Euler
] Problem 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. Find the largest palindrome made from the product of two 3-digit
·
2015-11-01 13:24
project
[Project
Euler
] Problem 3
The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? #include < iostream > using namespace std;
·
2015-11-01 13:24
project
[Project
Euler
] Problem 1
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.
·
2015-11-01 13:22
project
[Project
Euler
] 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 considering the
·
2015-11-01 13:22
project
[Project
Euler
]Problem 44
Pentagonal numbers are generated by the formula, Pn=n(3n1)/2. The first ten pentagonal numbers are: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ... It can be seen that P4 + P7 = 22 + 70 = 92 = P8. Howev
·
2015-11-01 12:00
project
[Project
Euler
]Problem 45
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae: Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal Pn=n(3n1)/2 1, 5, 12, 22, 35, ... Hexagonal Hn=n(2n1) 1, 6,
·
2015-11-01 12:00
project
[Project
Euler
]Problem 29
Consider all integer combinations of ab for 2 a 5 and 2 b 5: 2 2=4, 2 3=8, 2 4=16, 2 5=32 3 2=9, 3 3=27, 3 4=81, 3 5=243 4 2=16, 4 3=64, 4 4=256, 4 5=1024 5 2=25
·
2015-11-01 12:58
project
[Project
Euler
] Problem 21
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, then a and b are an amicable pair and each of a and b are
·
2015-11-01 12:57
project
[Project
Euler
] Problem 20
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! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the n
·
2015-11-01 12:57
project
UVa 10054 (打印欧拉回路) The Necklace
include <cstring> 3 4 const int maxn = 50 + 5; 5 int G[maxn][maxn], deg[maxn]; 6 7 void
Euler
·
2015-11-01 10:54
uva
[Project
Euler
] Problem 29
Consider all integer combinations of ab for 2 a 5 and 2 b 5: 2 2=4, 2 3=8, 2 4=16, 2 5=32 3 2=9, 3 3=27, 3 4=81, 3 5=243 4 2=16, 4 3=64, 4 4=256, 4 5=1024 5
·
2015-11-01 10:16
project
[Project
Euler
] Problem 27
Euler
published the remarkable quadratic formula: n² + n + 41 It turns out that the formula will
·
2015-11-01 10:15
project
[Project
Euler
] Problem 25
The Fibonacci sequence is defined by the recurrence relation: Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1. Hence the first 12 terms will be: F1 = 1 F2 = 1 F3 = 2 F4 = 3
·
2015-11-01 10:13
project
[Project
Euler
] Problem 26
这几日,一直忙于毕业设计,很久没有更新了 偶得闲暇,开始Level1的第一题 A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given: 1/2 = 0.5
·
2015-11-01 10:13
project
[Project
Euler
] Problem 24
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we c
·
2015-11-01 10:12
project
[Project
Euler
] Problem 23
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means th
·
2015-11-01 10:11
project
[Project
Euler
] Problem 22
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 alphabetical order. Then working out the alphabetical va
·
2015-11-01 10:10
project
HDOJ 2824 The
Euler
function (欧拉函数打表)
TheEulerfunctionTimeLimit:2000/1000MS(Java/Others) MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4658 AcceptedSubmission(s):1953ProblemDescriptionTheEulerfunctionphiisanimportantkindof
helloiamclh
·
2015-10-31 22:00
HDU 2824 The
Euler
function --------欧拉模板
The
Euler
function Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768
·
2015-10-31 19:21
function
[Project
Euler
] 来做欧拉项目练习题吧: 题目002
[Project
Euler
·
2015-10-31 15:47
project
[Project
Euler
] 来做欧拉项目练习题吧: 题目001
[Project
Euler
·
2015-10-31 15:46
project
混合图的欧拉回路
定义: 对于图G的一个回路,若它恰通过G中每条边一次,则称该回路为欧拉(
Euler
)回路。 具有欧拉回路的图称为欧拉图(简称E图)。
·
2015-10-31 15:17
图
hdu 2824 The
Euler
function(欧拉函数)
如果打表的话会超内存,我想到了一种方法解决这个问题。题目给出的数据时3000000,我将三百万分成300个数据,将整万的数据存储下来,计算的时候,先计算x和y之间整万的数据,然后再计算零散数据。 想法很不错,但为啥就是通不过呢?而且看提交返回的时间来看,应该是卡在最后一两组数据上了。 我很疑惑。 代码中注释的部分是我之前的代码。最后这道题竟然是毫无美感的暴力过去了。 #in
·
2015-10-31 14:19
function
Project
Euler
#80: Square root digital expansion
1 from decimal import getcontext, Decimal 2 3 4 def main(): 5 n = int(raw_input()) 6 p = int(raw_input()) 7 8 getcontext().prec = p+10 # 扩大精度,保证接过 9 sum = 0 10 1
·
2015-10-31 14:48
project
OGRE中的四元数与旋转
旋转的方式大致分为三种:
Euler
旋转,矩阵旋转,以及四元数旋转。 这里稍微记录下我目前对于四元数旋转的理解。对于四元数方面的数学,以及其原理,这里不关心,只需要学会如何使用即可。
·
2015-10-31 12:28
r
欧拉定理及其应用
Euler
定理:对于正整数a,p,若(a,p)=1,有aφ(p)mod p=1,其中欧拉函数φ(p)=p(1-1/p_1) (1-1/p_2)......(1-1/p_k),p_1,p_2,....,p_k
·
2015-10-31 11:44
应用
摄像机朝向指定物体
首先设置摄像机旋转角度,这里
Euler
(y, x, 0),意义为绕x轴旋转y角度,三维空间中绕x轴旋转即为上下旋转,绕y轴旋转x角度,三维空间中绕y轴旋转即为左右旋转。
·
2015-10-31 11:32
HDOJ---2824 The
Euler
function[欧拉函数]
The
Euler
function Time Limit: 2000/1000 MS (Java/Others) Memory Limit
·
2015-10-31 10:24
function
Quaternion 四元数运算
transform.rotation = Quaternion.
Euler
(new Vector3(45,a++,0));设置transform的旋转量为绕x轴旋转45度,y轴旋转量累加(y轴这时仍是世界坐标的旋转轴
·
2015-10-31 10:55
nio
POJ 3090 Visible Lattice Points 欧拉函数
include <iostream>#include <cmath>using namespace std; //点(x,y)可见当且仅当x,y互质,那么我仅仅只要知道int
euler
·
2015-10-31 10:57
visible
[Project
Euler
]加入欧拉 Problem 5
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
·
2015-10-31 09:38
project
[Project
Euler
]加入欧拉 Problem 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. Find the largest palindrome made from the product of two 3
·
2015-10-31 09:38
project
[Project
Euler
]加入欧拉 Problem 1
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. 这个
·
2015-10-31 09:38
project
[Project
Euler
]加入欧拉 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 considering t
·
2015-10-31 09:38
project
欧拉函数的简单求法及欧拉打表法
欧拉函数的意义为,对于一个正整数n来说,
euler
(n)即为小于等于n,且与n互斥的数的个数 根据容斥原理可得:
euler
(n)=n-n/p1-n/p2....
·
2015-10-31 09:02
函数
Euler
Circuit &
Euler
Trail
Trail: a walk that does not repeat any edges.Closed Trail(Circuit): a trail that begins and ends at the same vertex.Eulerian Circuit: a circuit that includes every edge of the graph.Eulerian Trail: a
·
2015-10-31 08:18
Euler
[Project
Euler
] Problem 31
In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possi
·
2015-10-31 08:45
project
UVA 10735
Euler
Circuit 混合图的欧拉回路(最大流,fluery算法)
题意:给一个图,图中有部分是向边,部分是无向边,要求判断是否存在欧拉回路,若存在,输出路径。 分析:欧拉回路的定义是,从某个点出发,每条边经过一次之后恰好回到出发点。 无向边同样只能走一次,只是不限制方向而已,那么这个情况下就不能拆边。不妨先按照所给的start和end的顺序,初步定下该无向边的顺序(若不当,一会再改)。那么有
·
2015-10-31 08:29
Euler
欧拉回路
欧拉通路、欧拉回路、欧拉图无向图:1) 设G是连通无向图,则称经过G的每条边一次并且仅一次的路径为欧拉通路;2) 如果欧拉通路是回路(起点和终点是同一个顶点),则称此回路为欧拉回路(
Euler
circuit
·
2015-10-31 08:25
That Nice
Euler
Circuit(LA3263+几何)
That Nice
Euler
Circuit Time Limit:3000MS Memory Limit:0KB
·
2015-10-30 18:22
Euler
[Project
Euler
] Problem 48
The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000. 用python解决这类问题实在是轻而易举了。一句话就可以了 print(str(sum([i**i for i
·
2015-10-30 15:35
project
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