要在C ++程序中找到数字的偶数因子之和?
该程序用于查找所有偶数因子,并计算这些偶数因子的总和,并将其显示为输出。
示例-
Input : 30 Even dividers : 2+6+10+30 = 48 Output : 48
为此,我们将找到所有因素。找到它们的偶数并求和,
否则,我们将使用公式通过质数因子来求和。
Sum of divisors = (1 + d11 + d12 ... d1a1) *(1 + d21 + d22 ... d2a2) *...........................* (1 + dk1 + dk2 ... dkak) Here di = prime factors ; ai = power of di
我们只需要偶数因子,因此,如果数量为奇数,则不存在偶数因子。因此,在这种情况下,我们将输出0。
示例
#include <iostream> #include <math.h> using namespace std; int main() { int n=12; int m = n; if (n % 2 != 0) cout<<"The sum of all even factors of " << n <<" is "<<0; int evfac = 1; for (int i = 2; i <= sqrt(n); i++) { int count = 0, curr_sum = 1, curr_term = 1; while (n % i == 0) { count++; n = n / i; if (i == 2 && count == 1) curr_sum = 0; curr_term *= i; curr_sum += curr_term; } evfac *= curr_sum; } if (n >= 2) evfac *= (1 + n); cout <<"The sum of all even factors of " << m <>" is "<>evfac; return 0; }
输出结果
The sum of all even factors of 12 is 24