?? ac1047.pas
字號(hào):
{Because a[i]=(a[i-1]+a[i+1])/2-c[i],
therefore 2*a[i]+2*c[i]=a[i-1]+a[i+1].
Let i be every integer between 1 and n inclusive, we have n formulae.
Adding all them together, we obtain
a[1]+a[n]+2*(c[1]+c[2]+...+c[n])=a[0]+a[n+1].
Denote c[1]+c[2]+...+c[n] by b[n],
then it becomes a[1]+a[n]+2*b[n]=a[0]+a[n+1].
Let n vary from 1 to n inclusive, we have another n formulae.
Adding all them together, we obtain
n*a[1]+a[1]+2*(b[1]+b[2]+...+b[n])=n*a[0]+a[n+1],
i.e.
a[1]=(n*a[0]+a[n+1]-2*(b[1]+b[2]+...+b[n]))/(n+1).
}
program ural1047;
var
n,i:integer;
a0,anp1,c,csum,bsum:real;
begin
read(n,a0,anp1);
csum:=0;bsum:=0;
for i:=1 to n do begin
read(c);
csum:=csum+c;
bsum:=bsum+csum;
end;
writeln((n*a0+anp1-2*bsum)/(n+1):0:2);
end.
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