交叉冲印滤镜又叫做正片负冲滤镜,关于它的原理我一直没找到,希望有大佬给我甩个链接. 我用的测试图像在这里 我实现的交叉冲印是使用上图这样的点运算来实现的,下面是我的代码 void punchFilter(Mat src) { int width = src.rows; int height = src.cols; int gray, value; int rMap[256], gMap[256], bMap[256]; for (int i = 0; i < 256; i++){ value =
那天的题挺简单的 下面来看下 No1 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. //project euler num1 #include <stdio.h> #inc
问题重述: 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 to 20? 问题的意思是找出可以整除小于等于某个数的所有公倍数,给出的测试用例是可以整除小