Int main ()
{
Int n, I;
Int A [20], [20] B, D [20];
Double C [20], sum [20].
The scanf (" % d ", & amp; N);
for(i=0; i
for(i=0; i
If (C [I]
=2)The sum [I]=5;
Else if (C [I] <10)
The sum [I]=5 + (int) ([I] - 2 (C)/0.5);
The else
The sum [I]=21 + (int) ((C [I] - 10)/0.5) * 1.5;
}
The else {
If (C [I]
=2)The sum [I]=6;
The else
The sum [I]=6 + (int) ([I] - 2 (C)/0.5) * 1.5;
}
[I] if (D & gt; 5)
The sum [I] +=[I] - 5 (D) * 0.5;
Printf (" % d \ n ", (int) (sum + 0.5) [I]);
}
return 0;
}
CodePudding user response:
Here I use devc++ is possibleCodePudding user response:
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