C++中的取余(rem)与取模(mod)

参考链接：https://blog.csdn.net/EiEddie23/article/details/122805453

#include<iostream>
#include<cstring>
#include<cmath>
#include <Windows.h>
using namespace std;

int mod(int x, int y)
{
    return x-y*int(int(double(x)/double(y)));
}

int rem(int x, int y)
{
    return x-y*int(floor(double(x)/double(y)));
}


int main()
{
    /*
     * floor是向负无穷取整
     * int是向0取整
     * 对于数学的取余来说floor
     * 对于c++的%来说是int
     * rem:
     * rem(5, 3) = 5-3*floor(5/3) = 5-3 = 2
     * rem(5, -3) = 5-(-3)*floor(5/-3) = 5-6 = -1
     * rem(-5, 3) = (-5)-3*floor(-5/3) = -5-3*(-2) = 1
     * rem(-5, -3) = (-5)-(-3)*floor(-5/-3) = (-5)-(-3) = -2
     *
     * mod(5, 3) = 5-3*int(5/3) = 5-3 = 2
     * mod(5, -3) = 5-(-3)*int(5/-3) = 5-3 = 2
     * mod(-5, 3) = (-5)-3*int(-5/3) = -5-3*(-1) = -2
     * mod(-5, -3) = (-5)-(-3)*int(-5/-3) = (-5)-(-3) = -2
     */
    cout<<"(5)%(3) "<<(5)%(3)<<endl;
    cout<<"(5)%(-3) "<<(5)%(-3)<<endl;
    cout<<"(-5)%(3) "<<(-5)%(3)<<endl;
    cout<<"(-5)%(-3) "<<(-5)%(-3)<<endl;
    cout<<endl;

    cout<<"mod(5, 3) "<<mod(5, 3)<<endl;
    cout<<"mod(5, -3) "<<mod(5, -3)<<endl;
    cout<<"mod(-5, 3) "<<mod(-5, 3)<<endl;
    cout<<"mod(-5, -3) "<<mod(-5, -3)<<endl;
    cout<<endl;

    cout<<"rem(5, 3) "<<rem(5, 3)<<endl;
    cout<<"rem(5, -3) "<<rem(5, -3)<<endl;
    cout<<"rem(-5, 3) "<<rem(-5, 3)<<endl;
    cout<<"rem(-5, -3) "<<rem(-5, -3)<<endl;
    return 0;
}

(5)%(3) 2
(5)%(-3) 2
(-5)%(3) -2
(-5)%(-3) -2

mod(5, 3) 2
mod(5, -3) 2
mod(-5, 3) -2
mod(-5, -3) -2

rem(5, 3) 2
rem(5, -3) -1
rem(-5, 3) 1
rem(-5, -3) -2