algorithm

问题 This question already has an answer here : Closed 7 years ago . Possible Duplicate: Efficient algorithm for conversion between numeral system Given an integer, write a program that converts the given number to a number (in base 10). Hint - The given number could be in any base, but the base is

问题 I extracted country outline data from somewhere and successfully managed to convert it into an array of lat-lng coordinates that I can feed to Google maps API to draw polyline or polygons. The problem is that that there are about 1200+ points in that shape. It renders perfectly in Google maps but

问题 What would be the best way to display the most faithfully graphics that have more than 8 bits per channel on a regular 24 bpp display? 回答1: The best solution I can think of is based on a random dithering that changes every frame. This combines the advantage of dithering with not having a fixed

问题 basically what i was trying to do is insert an integer k that represents the number of divisors and then finding all the numbers that have k divisors from 1-100000 #include <stdio.h> int main(void) { int k, x = 1, y = 100000, divisor, count; printf("Enter the target number of divisors:\n"); scanf

问题 Can someone tell me the Complexity of the Addition & Subtraction for the Divide & Conquer Matrix Multiplication algorithm? I know that the complexities of addition and subtraction operations of the Classic matrix multiplication are (n^3-n^2) while Strassen’s is 6n^2.81 – 6n^2... but I can't seem

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 4 years ago . let's say we have to print:(for n=4) 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7 this is just an sample

问题 int lcs(char * A, char * B) { int m = strlen(A); int n = strlen(B); int *X = malloc(m * sizeof(int)); int *Y = malloc(n * sizeof(int)); int i; int j; for (i = m; i >= 0; i--) { for (j = n; j >= 0; j--) { if (A[i] == '\0' || B[j] == '\0') X[j] = 0; else if (A[i] == B[j]) X[j] = 1 + Y[j+1]; else X

问题 I was writing a very simple program to examine if a number could divide another number evenly: // use the divider squared to reduce iterations for(divider = 2; (divider * divider) <= number; divider++) if(number % divider == 0) print("%d can divided by %d\n", number, divider); Now I was curious

问题 So I N - numbers I have to input, and I got M - numbers of places for those numbers and I need to find all combinations with repetition of given numbers. Here is example: Let's say that N is 3(I Have to input 3 numbers), and M is 4. For example let's input numbers: 6 11 and 533. This should be

问题 Given an array with integer 0 to N, how many ways to arrange it such that at position i of the array, we cannot have i inserted in it? For example, N = 2 The following arrangements is valid: 1,2,0 2,0,1 Thus, the answer is 2 arrangements I can't think of a non-brute force method to do this in O(1