Highly divisible triangular number 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, 36,…

我的那个暴力求解,太耗时间了. 用了网上产的什么因式分解,质因数之类的.确实快!还是数学基础不行,只能知道大约. 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, 2…

最直接的想法就是暴力搞搞,直接枚举,暴力分解因子.再好一点,就打个素数表来分解因子.假设num=p1^a1p2^a2...pn^an,则所有因子个数为(a1+1)(a2+1)...(an+1). 再好一点呢,还是要利用积性函数.假设因子个数函数为f(x),因为num=n(n+1)/2,所以num2=n(n+1),因为n和n+1为相邻的正整数,所以互质,所以f(2num)=f(n)f(n+1), 则f(num)=f(n)f((n+1)/2)(n+1为偶数)或者f(num)=f(n/2)*f(n+1…

title: 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, 36, 45, 55, ... Let us list the factors of…

Summation of primes 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. The code resemble : import math limit = 2000000 crosslimit = int(math.sqrt(limit)) #sieve = [False] * limit sieve =…

题意须要注意的一点就是, 序列是从外层最小的那个位置顺时针转一圈得来的.而且要求10在内圈 所以,这题暴力的话,假定最上面那个点一定是第一个点,算下和更新下即可. #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cstdlib> #include <ctime> #include <set> #i…

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 100 (计算1000以内能被3或5整除的数之和) #include <iostream> using namespace std; i…

直接暴力 写完才想到,代码写矬了,扫一遍就出结果了,哪还用我写的这么麻烦 ss = '''73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 6…

题目的大意很简单 做法的话. 我们就枚举1~10000的数的立方, 然后把这个立方中的有序重新排列,生成一个字符串s, 然后对于那些符合题目要求的肯定是生成同一个字符串的. 然后就可以用map来搞了 这里偷懒用了python import string dic = {} def getstr(n): sn = str(n) arr = [] for c in sn: arr.append(c) arr.sort() return ''.join(arr) for i in xrange(1, 1…

这题略水啊 首先观察一下. 10 ^ x次方肯定是x + 1位的 所以底数肯定小于10的 那么我们就枚举1~9为底数 然后枚举幂级数就行了,直至不满足题目中的条件即可break cnt = 0 for i in range(1, 10): e = 1 while True: if len(str(i**e)) != e: break e += 1 cnt += 1 print cnt…