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

#include<bits/stdc++.h> using namespace std; int main(){ cout<<"4396"<<"+"<<"2200"<<"="<<"6596"<<endl; cout<<"4396"<<"-"<<"2200"<<"="<<"2196"<<endl; cout<<"4396"<<"*"<<"2200"<<"="<<"9671200"<<endl; cout<<"4396"<<"/"<<"2200"<<"="<<4396/2200<<endl; cout<<"4396"<<"%"<<"2200"<<"="<<"2196"; return 0; }

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

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

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

#include <bits/stdc++.h> using namespace std; int main() { cout<<"4396+2200="<<2200+4396<<endl; cout<<"4396-2200="<<4396-2200<<endl; cout<<"43962200="<<22004396<<endl; cout<<"4396/2200="<<4396/2200<<endl; cout<<"4396%2200="<<4396%2200<<endl; }

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

典型的求和差积商余题目，我省事，直接算出了结果：

#include <bits/stdc++.h> using namespace std; int main(){ int a=2200,b=4396; cout <<b<<'+'<<a<<'='<<a+b<<endl; cout <<b<<'-'<<a<<'='<<b-a<<endl; cout <<b<<''<<a<<'='<<ba<<endl; cout <<b<<'/'<<a<<'='<<b/a<<endl; cout <<b<<'%'<<a<<'='<<b%a; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int h=2200,c=4396; cout<<c<<"+"<<h<<"="<<c+h<<endl; cout<<c<<"-"<<h<<"="<<c-h<<endl; cout<<c<<""<<h<<"="<<ch<<endl; cout<<c<<"/"<<h<<"="<<c/h<<endl; cout<<c<<"%"<<h<<"="<<c%h;//证明：我没复制！！！ //不写return 0；好习惯！！！ }

众所周知，这种简简单单的计算题是一年级的小朋友们都可以口算的。 所以，废话不多说，上最少内存的题解！ 等等 点击加入“ACGO的AC团”

#include<bits/stdc++.h> using namespace std; int n=4396,m=2200; int main(){ cout<<n<<"+"<<m<<"="<<n+m<<"\n"; cout<<n<<"-"<<m<<"="<<n-m<<"\n"; cout<<n<<""<<m<<"="<<nm<<"\n"; cout<<n<<"/"<<m<<"="<<n/m<<"\n"; cout<<n<<"%"<<m<<"="<<n%m<<"\n"; return 0; }