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) −
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_countend
find_div(a, b) –
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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 doneend
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 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
以上就是在C/C++中编写求解模方程的程序？的详细内容，更多请关注创想鸟其它相关文章！
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