Circulation structure in the application of piecewise function how to evaluate value of y?-CodePudding

Known x keyboard input by the user, calculate the following section function y value
Y=x2 + 10 (x <5)
Y=(x + 1) * (x + 5) <=(- 5 x <0)
Y=x (x=0)
Y=x - 2 (0 Y=x + 25 (x>=10)
Out, I want to use the if the else structure found that logic is wrong, who can type the complete code, I compare,

CodePudding user response:

If you can, take explanation, by the way

CodePudding user response:

Reference:

 # include 
Int main (int arg c, char * argv []) 
{
Float x, y; 
Printf (" input x: "); 
The scanf (" % f ", & amp; X); 
If (x<5) 
{
X + y=x * 10; 
Printf (" y=% f \ n ", y); 
} 
If (x & gt;=5 & amp; & X & lt; 0) 
{
Y=(x + 1) * (x + 5) + 0.000000001;//to avoid - 0.00 
 the condition of thePrintf (" y=% f \ n ", y); 
} 
If (x==0) 
{
Y=x; 
Printf (" y=% f \ n ", y); 
} 
If (x> 0 & amp; & x<10) 
{
Y=x - 2; 
Printf (" y=% f \ n ", y); 
} 
If (x>=10) 
{
Y=x + 25; 
Printf (" y=% f \ n ", y); 
} 
system("pause"); 
return 0; 
}

CodePudding user response:

refer to the second floor QZJHJXJ response:

for reference:

 # include 
Int main (int arg c, char * argv []) 
{
Float x, y; 
Printf (" input x: "); 
The scanf (" % f ", & amp; X); 
If (x<5) 
{
X + y=x * 10; 
Printf (" y=% f \ n ", y); 
} 
If (x & gt;=5 & amp; & X & lt; 0) 
{
Y=(x + 1) * (x + 5) + 0.000000001;//to avoid - 0.00 
 the condition of thePrintf (" y=% f \ n ", y); 
} 
If (x==0) 
{
Y=x; 
Printf (" y=% f \ n ", y); 
} 
If (x> 0 & amp; & x<10) 
{
Y=x - 2; 
Printf (" y=% f \ n ", y); 
} 
If (x>=10) 
{
Y=x + 25; 
Printf (" y=% f \ n ", y); 
} 
system("pause"); 
return 0; 
}

Thank you, after I find the wrong place