POJ 2456(二分哲学）
这题普通的二分会T…………
法一：只循环60遍，用ans记录答案(见标程）
法二：进行特判，若l+1==r 则 m=(l+r+1) shl 1 否则 m=(l+r) shl 1
Program P2456;const maxd=1000000000; maxn=100000;var n,c,i,j,k:longint; a:array[1..maxn] of longint;procedure sort(l,r:longint);var m,i,k,head,tot,ans:longint;begin for k:=1 to 60 do begin m:=(l+r) shr 1; head:=1;tot:=0; for i:=2 to n do begin if a[i]-a[head]>=m then begin inc(tot); head:=i; end; end; if tot>=c-1 then begin ans:=m; l:=m+1 end else r:=m-1; end; writeln(ans);end;procedure qsort(l,r:longint);var i,j,m,p:longint;begin i:=l; j:=r; m:=a[(l+r) shr 1]; repeat while a[i]<m do inc(i); while a[j]>m do dec(j); if i<=j then begin p:=a[i];a[i]:=a[j];a[j]:=p; inc(i);dec(j); end; until i>j; if l<j then qsort(l,j); if i<r then qsort(i,r);end;begin read(n,c); for i:=1 to n do read(a[i]); qsort(1,n); sort(1,maxd);end.
