#include<iostream> using namespace std; int main(){ cout<<"4396+2200="<<4396+2200<<endl; cout<<"4396-2200="<<4396-2200<<endl; cout<<"43962200="<<43962200<<endl; cout<<"4396/2200="<<4396/2200<<endl; cout<<"4396%2200="<<4396%2200; return 0; }

#include <iostream> using namespace std; int baron(int h,int g){ cout<<h<<"+"<<g<<"="<<h+g<<endl; cout<<h<<"-"<<g<<"="<<h-g<<endl; cout<<h<<""<<g<<"="<<hg<<endl; cout<<h<<"/"<<g<<"="<<h/g<<endl; cout<<h<<"%"<<g<<"="<<h%g<<endl; } int main() { int h=4396,g=2200; baron(h,g); return 0; }

#include <iostream> using namespace std; int baron(long long h,long long g){ cout<<h<<"+"<<g<<"="<<h+g<<endl; cout<<h<<"-"<<g<<"="<<h-g<<endl; cout<<h<<""<<g<<"="<<hg<<endl; cout<<h<<"/"<<g<<"="<<h/g<<endl; cout<<h<<"%"<<g<<"="<<h%g<<endl; } int main() { int h=4396,g=2200; baron(h,g); return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int h=4396; int g=2200; cout<<h<<"+"<<g<<"="<<h+g<<endl; cout<<h<<"-"<<g<<"="<<h-g<<endl; cout<<h<<""<<g<<"="<<hg<<endl; cout<<h<<"/"<<g<<"="<<h/g<<endl; cout<<h<<"%"<<g<<"="<<h%g<<endl; return 0; }

我的才是对的，给个赞就行。 #include <iostream> using namespace std; int main(){ int g = 2200,h = 4396; cout << h << "+" << g << "=" << h + g << endl; cout << h << "-" << g << "=" << h - g << endl; cout << h << "*" << g << "=" << h * g << endl; cout << h << "/" << g << "=" << h / g << endl; cout << h << "%" << g << "=" << h % g; return 0; }

只给了核心代码，其他请自主编写！

在这个代码中只不过用了很多的#define和一个函数void _Output

#include <iostream> using namespace std; int main() { int h=4396,g=2200; cout<<h<<"+"<<g<<"="<<h+g<<endl; cout<<h<<"-"<<g<<"="<<h-g<<endl; cout<<h<<""<<g<<"="<<hg<<endl; cout<<h<<"/"<<g<<"="<<h/g<<endl; cout<<h<<"%"<<g<<"="<<h%g<<endl; return 0; }

#include <iostream> using namespace std; int main() { int g=4396,h=2200; cout<<"4396+2200="<<g+h<<endl; cout<<"4396-2200="<<g-h<<endl; cout<<"43962200="<<gh<<endl; cout<<"4396/2200="<<g/h<<endl; cout<<"4396%2200="<<g%h<<endl; return 0; }

A730.表达式计算2 难度：入门 来源：官方 ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ AC代码（C++/C#）： 没有注释，自行学习。 by：ጿ ኈ ቼ ዽ ጿ ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 你的点赞和关注就是我更新题解的最大动力！ ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ -----------------------------点个赞吧 ↓ ！-----------------------------

#include <iostream> using namespace std; int main() { int h = 4396, g = 2200; printf("%d+%d=%d\n",h,g,h+g); printf("%d-%d=%d\n",h,g,h-g); printf("%d*%d=%d\n",h,g,h*g); printf("%d/%d=%d\n",h,g,h/g); printf("%d",h); cout<<"%"; printf("%d=%d\n",g,h%g); return 0; }

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 main(){ cout<<"4396+2200="<<4396+2200<<endl; cout<<"4396-2200="<<4396-2200<<endl; cout<<"43962200="<<43962200<<endl; cout<<"4396/2200="<<4396/2200<<endl; cout<<"4396%2200="<<4396%2200<<endl; return 0; }