[英]How to transform this code to work for more than two numbers (calculating HCF)
I don't know how to make this code to work for more than two numbers. 我不知道如何使这个代码适用于两个以上的数字。 It works perfectly for two numbers.
它适用于两个数字。
#include <iostream>
using namespace std;
int main() {
int n1, n2;
cout << " Insert 2 numbers: ";
cin >> n1 >> n2;
while(n1 != n2)
{
if(n1 > n2)
{
n1 -= n2;
}
else
{
n2 -= n1;
}
}
i cout << "HCF = " << n1; return 0;
}
For example if we input 6 and 12 the code says 6 which is right. 例如,如果我们输入6和12,则代码表示6是正确的。
Just turn your calculation into a function: 只需将计算转换为函数:
#include <iostream>
int hcf(int n1, int n2);
int main() {
int n1, n2, n3;
std::cout << " Insert 3 numbers: ";
std::cin >> n1 >> n2 >> n3;
std::cout << "HCF = " << hcf(n1, hcf(n2, n3));
return 0;
}
int hcf(int n1, int n2) {
while (n1 != n2) {
if (n1 > n2)
n1 -= n2;
else
n2 -= n1;
}
return n1;
}
Now you can easily calculate HCF for as many numbers as you want. 现在,您可以根据需要轻松计算HCF数量。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.