bzoj2038

网上大片的莫队算法题解，先orz一下莫队
什么不会莫队？没事我来篇低端的
分块大法好啊，我们知道对于区间[l,r]答案是S/P P是一下子可以算出来的，S=∑（pj-1)*pj/2 pj表示区间内颜色为j的数目
先记录ans[i,j]表示第i块到第j块之间的S
然后再记录颜色g[i,j]表示到第i块颜色为j的数目
然后暴力统计不在整块内颜色对答案的贡献即可
20分钟就1A了，非常好写
虽然理论复杂度和莫队差不多O(nsqrt(n))
但是实际上似乎被莫队完败了，但是没关系，毕竟我们这是在线算法啊

var f:array[0..230,0..230] of longint;
    g:array[0..50010,0..230] of longint;
    be,c,s:array[0..50010] of longint;
    size,p,t,i,n,m,x,y:longint;

function gcd(a,b:longint):longint;
  begin
    if b=0 then exit(a)
    else exit(gcd(b,a mod b));
  end;

procedure prework;
  var i,j:longint;
  begin
    for i:=1 to n do
      inc(g[c[i],be[i]]);
    for i:=1 to p do
      for j:=2 to t do
        inc(g[i,j],g[i,j-1]);
    for i:=1 to t do
    begin
      for j:=(i-1)*size+1 to n do
      begin
        if j mod size=1 then f[i,be[j]]:=f[i,be[j]-1];
        inc(f[i,be[j]],s[c[j]]);
        inc(s[c[j]]);
      end;
      fillchar(s,sizeof(s),0);
    end;
  end;

procedure clear(l,r:longint);
  var i:longint;
  begin
    for i:=l to r do
      s[c[i]]:=0;
  end;

procedure ask(x,y:longint);
  var i,a,b,d:longint;
  begin
    b:=int64(y-x+1)*int64(y-x) div 2;
    a:=0;
    if be[x]=be[y] then
    begin
      for i:=x to y do
      begin
        a:=a+s[c[i]];
        inc(s[c[i]]);
      end;
      clear(x,y);
    end
    else begin
      a:=f[be[x]+1,be[y]-1];
      for i:=x to be[x]*size do
      begin
        if s[c[i]]=0 then s[c[i]]:=g[c[i],be[y]-1]-g[c[i],be[x]];
        a:=a+s[c[i]];
        inc(s[c[i]]);
      end;
      for i:=(be[y]-1)*size+1 to y do
      begin
        if s[c[i]]=0 then s[c[i]]:=g[c[i],be[y]-1]-g[c[i],be[x]];
        a:=a+s[c[i]];
        inc(s[c[i]]);
      end;
      clear(x,be[x]*size);
      clear((be[y]-1)*size+1,y);
    end;
    d:=gcd(a,b);
    writeln(a div d,'/',b div d);
  end;

begin
  readln(n,m);
  size:=trunc(sqrt(n));
  for i:=1 to n do
  begin
    read(c[i]);
    if c[i]>p then p:=c[i];
    be[i]:=(i-1) div size+1;
  end;
  t:=i div size;
  if i mod size<>0 then inc(t);
  prework;
  for i:=1 to m do
  begin
    readln(x,y);
    ask(x,y);
  end;
end.