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我想說(shuō)不考慮算法選擇要求 這就是個(gè) 2x+b = a 求x的小學(xué)數(shù)學(xué)題，如果對(duì)算法選擇沒(méi)要求就是輸入a,b,輸出(a-b)/2 和 （a+b）/ 2.
輸入的時(shí)候校驗(yàn)下a,b是不是同為奇數(shù)或偶數(shù)以及數(shù)字是不是超變量類型的可用范圍，然后自己生成的case也要考慮這個(gè)問(wèn)題

雷雷

通過(guò)你提供的材料,用c++編碼如下,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;
}

運(yùn)行結(jié)果