3、面向过程的C实现
这是 csdn 算法论坛前版主海星的代码,程序非常简练、精致:
#include #include #include using namespace std; const double PRECISION = 1E-6; const int COUNT_OF_NUMBER = 4; const int NUMBER_TO_BE_CAL = 24; double number[COUNT_OF_NUMBER]; string expression[COUNT_OF_NUMBER]; bool Search(int n) { if (n == 1) { if ( fabs(number[0] - NUMBER_TO_BE_CAL) < PRECISION ) { cout << expression[0] << endl; return true; } else { return false; } } for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { double a, b; string expa, expb; a = number[i]; b = number[j]; number[j] = number[n - 1]; expa = expression[i]; expb = expression[j]; expression[j] = expression[n - 1]; expression[i] = '(' + expa + '+' + expb + ')'; number[i] = a + b; if ( Search(n - 1) ) return true; expression[i] = '(' + expa + '-' + expb + ')'; number[i] = a - b; if ( Search(n - 1) ) return true; expression[i] = '(' + expb + '-' + expa + ')'; number[i] = b - a; if ( Search(n - 1) ) return true; expression[i] = '(' + expa + '*' + expb + ')'; number[i] = a * b; if ( Search(n - 1) ) return true; if (b != 0) { expression[i] = '(' + expa + '/' + expb + ')'; number[i] = a / b; if ( Search(n - 1) ) return true; } if (a != 0) { expression[i] = '(' + expb + '/' + expa + ')'; number[i] = b / a; if ( Search(n - 1) ) return true; } number[i] = a; number[j] = b; expression[i] = expa; expression[j] = expb; } } return false; } void main() { for (int i = 0; i < COUNT_OF_NUMBER; i++) { char buffer[20]; int x; cin >> x; number[i] = x; itoa(x, buffer, 10); expression[i] = buffer; } if ( Search(COUNT_OF_NUMBER) ) { cout << "Success." << endl; } else { cout << "Fail." << endl; } } |
使用任一个 c++ 编译器编译即可。 |