入 门 题 解

用了两个函数而已
有没有人能把它更加*山一点啊

#include <iostream>
using namespace std;

int f1(int h, int g, char c){
    if (c == '+') return h + g;
    else if (c == '-') return h - g;
    else if (c == '*') return h * g;
    else if (c == '/') return h / g;
    else return h % g;
}

void f2(){
    cout << "4396+2200=" << f1(4396, 2200, '+') << endl;
    cout << "4396-2200=" << f1(4396, 2200, '-') << endl;
    cout << "4396*2200=" << f1(4396, 2200, '*') << endl;
    cout << "4396/2200=" << f1(4396, 2200, '/') << endl;
    cout << "4396%2200=" << f1(4396, 2200, '%');
}

int main(){
    f2();
    return 0;
}