number-theory

问题 Consider the finite set {2,3,5,...,n}. I am interested in primes but the question could apply to any set of numbers. I want to find all possible products of these numbers in ascending order, and in particular greater than or equal to some number x. Does anyone know a nice algorithm for this? EDIT to clarify: Each factor in the input set may be used any number of times. If the input were {2,3,5,7} the output would be {2,3,4,5,6,7,8,9,10,12,14,15,16,18,...}. The algorithm can stop as soon as it

问题 I need a way to align tick marks on two separate axis, while being able to control the "step" value (value between tick marks), where both axis start at mark 0 and end on a different maximum value. Why this problem: Flot, the JS charting package has an option to align tick marks, but when I do, I cannot control the step value. I can however control the step value directly, but then I lose the ability to align tick marks. I can however revert to defining my own max and step values, to get what

问题 Given a number N, have to find number the divisors for all i where i>=1 and i<=N. Can't figure it out.Do I have to this using prime factorization? Limit is N<=10^9 Sample Output: 1 --> 1 2 --> 3 3 --> 5 4 --> 8 5 --> 10 6 --> 14 7 --> 16 8 --> 20 9 --> 23 10 --> 27 11 --> 29 12 --> 35 13 --> 37 14 --> 41 15 --> 45 回答1: To compute much faster, use the following Pseudocode: sum = 0; u = floor(sqrt(N)); foreach k <= u, sum += Floor(N / K); sum = 2*sum-u*u The above formula is given by Peter

问题 Here is the problem (Summation of Four Primes) states that : The input contains one integer number N (N<=10000000) in every line. This is the number you will have to express as a summation of four primes Sample Input: 24 36 46 Sample Output: 3 11 3 7 3 7 13 13 11 11 17 7 This idea comes to my mind at a first glance Find all primes below N Find length of list (.length = 4) with Integer Partition problem (Knapsack) but complexity is very bad for this algorithm I think. This problem also looks

问题 The code below works perfectly but I would like someone to explain to me the mathematics behind it. Basically, how does it work? #include <stdio.h> #include <stdlib.h> /* atoi */ #define min(x, y) (((x) < (y)) ? (x) : (y)) int main(int argc, char* argv[]) { const int base = 16; int n,i,j,p,c,noz,k; n = 7; /* 7! = decimal 5040 or 0x13B0 - 1 trailing zero */ noz = n; j = base; /* Why do we start from 2 */ for (i=2; i <= base; i++) { if (j % i == 0) { p = 0; /* What is p? */ while (j % i== 0) {

问题 f(N) = 0^0 + 1^1 + 2^2 + 3^3 + 4^4 + ... + N^N. I want to calculate ( f(N) mod M ). These are the constraints. 1 ≤ N ≤ 10^9 1 ≤ M ≤ 10^3 Here is my code test=int(input()) ans = 0 for cases in range(test): arr=[int(x) for x in input().split()] N=arr[0] mod=arr[1] #ret=sum([int(y**y) for y in range(N+1)]) #ans=ret for i in range(1,N+1): ans = (ans + pow(i,i,mod))%mod print (ans) I tried another approach but in vain. Here is code for that from functools import reduce test=int(input()) answer=0

问题 Is there any algorithm to calculate (1^x + 2^x + 3^x + ... + n^x) mod 1000000007 ? Note: a^b is the b-th power of a. The constraints are 1 <= n <= 10^16, 1 <= x <= 1000 . So the value of N is very large. I can only solve for O(m log m) if m = 1000000007 . It is very slow because the time limit is 2 secs. Do you have any efficient algorithm? There was a comment that it might be duplicate of this question, but it is definitely different. 回答1: You can sum up the series 1**X + 2**X + ... + N**X

问题 I have to compute efficiently a^^b mod m for large values of a,b,m<2^32 where ^^ is the tetration operator: 2^^4=2^(2^(2^2)) m is not a prime number and not a power of ten. Can you help? 回答1: To be clear, a^^b is not the same thing as a^b, it is the exponential tower a^(a^(a^...^a)) where there are b copies of a, also known as tetration. Let T(a,b) = a^^b so T(a,1) = a and T(a,b) = a^T(a,b-1). To compute T(a,b) mod m = a^T(a,b-1) mod m, we want to compute a power of a mod m with an extremely

问题 I'd like to take the modular inverse of a matrix like [[1,2],[3,4]] mod 7 in Python. I've looked at numpy (which does matrix inversion but not modular matrix inversion) and I saw a few number theory packages online, but nothing that seems to do this relatively common procedure (at least, it seems relatively common to me). By the way, the inverse of the above matrix is [[5,1],[5,3]] (mod 7). I'd like Python to do it for me though. 回答1: A hackish trick which works when rounding errors aren't an

问题 I have a series S = i^(m) + i^(2m) + ............... + i^(km) (mod m) 0 <= i < m, k may be very large (up to 100,000,000), m <= 300000 I want to find the sum. I cannot apply the Geometric Progression (GP) formula because then result will have denominator and then I will have to find modular inverse which may not exist (if the denominator and m are not coprime). So I made an alternate algorithm making an assumption that these powers will make a cycle of length much smaller than k (because it