我想说不考虑算法选择要求 这就是个 2x+b = a 求x的小学数学题，如果对算法选择没要求就是输入a,b,输出(a-b)/2 和 （a+b）/ 2.
输入的时候校验下a,b是不是同为奇数或偶数以及数字是不是超变量类型的可用范围，然后自己生成的case也要考虑这个问题

雷雷

通过你提供的材料,用c++编码如下,望指正!

/*
  if Natalia's number of apples is x  (x>0)
  apples:the number of apple
  extra:the number that Klaudia more tha Natalia
  (apples>extra>0)
  so
  x+extra+x=apples
  x = (apples-extra)/2
*/
#include <iostream>
int main(int argc, char const *argv[]) {
  int apples,extra;   //apples:the number of apple  extra:the number that Klaudia more tha Natalia

  for (size_t i = 0; i < 10; i++) {
        std::cin>>apples>>extra;   //input
        try{
          //To determine whether the input is legal
            if(apples<=0 || extra<=0 || apples<extra || (apples-extra)%2!=0) throw apples;
            int Klaudia,Natalia;
            Natalia = (apples-extra)/2;    //calculate
            Klaudia=Natalia+2;
            std::cout<<Klaudia<<'\n'<<Natalia<<'\n';    //output
      }
        catch(int e){
            //ERROR
            std::cerr << "ERROR! the number of applse is error.\n" << '\n';
        }
  }
  return 0;
}

运行结果