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) -
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 <iostream>#include <cmath>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
以上就是在c/c++中编写求解模方程的程序？的详细内容。