CPP5. 简单运算
描述
键盘输入两个 int 范围的正整数 a 和 b,范围为[1, 9999],输出这两个 int 范围的正整数的和,差,积,商,模(若 a > b 则输出 a - b,a / b,a % b 的值反之输出 b - a,b / a,b % a 的值,不考虑小数)输入描述
两个 int 范围的正整数,范围为[1, 9999]输出描述
输出这两个 int 范围的正整数的和,差,积,商,模(若 a > b 则输出 a - b,a / b,a % b 的值反之输出 b - a,b / a,b % a 的值,不考虑小数和数据越界情况)示例1
输入:
10 5
输出:
15 5 50 2 0
C++ 解法, 执行用时: 2ms, 内存消耗: 296KB, 提交时间: 2022-08-06
#include <iostream> using namespace std; int main() { // write your code here...... int a,b,sum,chai,ji,shang,mo; cin>>a; cin>>b; if(a>b){ sum=a+b; chai=a-b; ji=a*b; shang=a/b; mo=a%b; cout<<sum<<" "<<chai<<" "<<ji<<" "<<shang<<" "<<mo<<endl; } else{ sum=a+b; chai=b-a; ji=a*b; shang=b/a; mo=b%a; cout<<sum<<" "<<chai<<" "<<ji<<" "<<shang<<" "<<mo<<endl; } return 0; }
C++ 解法, 执行用时: 2ms, 内存消耗: 312KB, 提交时间: 2021-11-21
#include <iostream> using namespace std; int main() { // write your code here...... int a , b ; cin >> a >> b; if(a < 1&& a > 9999 || b < 1&& b > 9999){ cout << "erorr\n" << endl; }else{ if(a > b){ cout << a + b << " " << a - b << " " << a * b << " " << a / b << " " << a % b << endl; }else{ cout << a + b << " " << b - a << " " << a * b << " " << b / a << " " << b % a << endl; } } return 0; }
C++ 解法, 执行用时: 2ms, 内存消耗: 332KB, 提交时间: 2022-07-18
#include <iostream> using namespace std; int main() { int a,b; cin>>a>>b; if(a>=1&&a<=9999&&b>=1&&b<=9999){ if(a>b) cout<<a+b<<" "<<a-b<<" "<<a*b<<" "<<a/b<<" "<<a%b<<endl; else cout<<a+b<<" "<<b-a<<" "<<a*b<<" "<<b/a<<" "<<b%a<<endl; } // write your code here...... return 0; }
C++ 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2022-01-27
#include <iostream> using namespace std; int main() { int a,b; cin>>a>>b; if(a>b){ cout<<a+b<<" "<<a-b<<" "<<a*b<<" "<<a/b<<" "<<a%b;} else{ cout<<a+b<<" "<<b-a<<" "<<a*b<<" "<<b/a<<" "<<b%a;} // write your code here...... return 0; }
C++ 解法, 执行用时: 2ms, 内存消耗: 384KB, 提交时间: 2022-01-26
#include <iostream> using namespace std; int main() { // write your code here...... int a, b; cin >> a >> b; if(a < b) //交换,使a比b大 swap(a, b); //输出时计算 cout << a + b << " " << a - b << " " << a * b << " " << a / b << " "<< a % b << endl; return 0; }