1 solutions

  • 0
    @ 2025-11-5 19:44:15

    C :

    int main(int argc, char* argv[])
    {
    	int a,b;
    	while(~scanf("%d%d",&a,&b))
    	{
    	 printf("%.1lf\n",(5-3*b)*1.0/(2*a));
    	}
    	return 0;
    }
    
    

    C++ :

    #include <iostream>
    #include<iomanip>
    using namespace std;
    
    int main()
    {
        int a,b;
        double x;
        cin>>a>>b;
        x=(5-3*b)*1.0/(2*a);
        cout<<fixed<<setprecision(1);
        cout<<x<<endl;
        return 0;
    }
    
    
    • 1

    《C语言程序设计》江宝钏主编-习题1-6-解方程

    Information

    ID
    19610
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    (None)
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