在C/C++中编写求解模方程的程序？-C++-PHP中文网

在c/c++中编写求解模方程的程序？

在这里，我们将看到一个与模方程相关的有趣问题。假设我们有两个值A和B。我们必须找到变量X可以取的可能值的数量，使得(A mod X) = B成立。

假设A为26，B为2。所以X的首选值将是{3, 4, 6, 8, 12, 24}，因此计数为6。这就是答案。让我们看一下算法以更好地理解。

算法

possibleWayCount(a, b) −

云网OA

采用JSP开发的办公自动化产品、基于B/S结构，运行环境：JDK v1.5、Tomcat v5.5、MySQL v4.1，三者均为以上版本其他相关内容：可视化流程设计：流程支持串签、会签和分支流程，可以设置流程节点的修改、删除权限，并可指定流程中各个用户在表单中可以填写的域。智能表单所见即所得设计：智能设计，自动在数据库中生成表格，方便优化程序公共交流：集论坛、博客、聊天室于一体文件柜：C

下载

begin
   if a = b, then there are infinite solutions
   if a < b, then there are no solutions
   otherwise div_count := find_div(a, b)
   return div_count
end

find_div(a, b) -

立即学习“C++免费学习笔记（深入）”；

begin
   n := a – b
   div_count := 0
   for i in range 1 to square root of n, do
      if n mode i is 0, then
         if i > b, then
            increase div_count by 1
         end if
         if n / i is not same as i and (n / i) > b, then
            increase div_count by 1
         end if
      end if
   done
end

Example

的中文翻译为：

示例

#include 
#include 
using namespace std;
int findDivisors(int A, int B) {
   int N = (A - B);
   int div_count = 0;
   for (int i = 1; i <= sqrt(N); i++) {
      if ((N % i) == 0) {
         if (i > B)
            div_count++;
         if ((N / i) != i && (N / i) > B) //ignore if it is already counted
            div_count++;
      }
   }
   return div_count;
}
int possibleWayCount(int A, int B) {
   if (A == B) //if they are same, there are infinity solutions
      return -1;
   if (A < B) //if A < B, then there are two possible solutions
      return 0;
   int div_count = 0;
   div_count = findDivisors(A, B);
   return div_count;
}
void possibleWay(int A, int B) {
   int sol = possibleWayCount(A, B);
   if (sol == -1)
      cout << "For A: " << A << " and B: " << B << ", X can take infinite values greater than " << A;
   else
      cout << "For A: " << A << " and B: " << B << ", X can take " << sol << " values";
}
int main() {
   int A = 26, B = 2;
   possibleWay(A, B);
}

输出

For A: 26 and B: 2, X can take 6 values