【ZJOI2017 Round1练习&BZOJ4767】D1T3 两双手（排列组合，DP）

题意：

100%的数据：|Ax|,|Ay|,|Bx|,|By| <= 500, 0 <= n,Ex,Ey <= 500

思路：听说这是一道原题

只能往右或者下走一步且有禁止点的简化版是CF559C

然而这道题并没有这么简单

以下开始转化：

转化后套用弱化版做法即可

 const mo=;
 var fac,exf:array[..]of int64;
     dp:array[..]of int64;
     x,y:array[..]of longint;
     ex,ey,sx,sy,ax,ay,bx,by:int64;
     n,n1,i,j:longint;
     u,v,eps:double;

 function fabs(x:double):double;
 begin
  if x< then exit(-x);
  exit(x);
 end;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 procedure qsort(l,r:longint);
 var i,j,mid1,mid2:longint;
 begin
  i:=l; j:=r; mid1:=x[(l+r)>>]; mid2:=y[(l+r)>>];
  repeat
   while (mid1>x[i])or((mid1=x[i])and(mid2>y[i])) do inc(i);
   while (mid1<x[j])or((mid1=x[j])and(mid2<y[j])) do dec(j);
   if i<=j then
   begin
    swap(x[i],x[j]);
    swap(y[i],y[j]);
    inc(i); dec(j);
   end;
  until i>j;
  if l<j then qsort(l,j);
  if i<r then qsort(i,r);
 end;

 function c(x,y:longint):int64;
 begin
  exit(fac[x]*exf[y] mod mo*exf[x-y] mod mo);
 end;

 begin
  assign(input,'hands.in'); reset(input);
  assign(output,'hands.out'); rewrite(output);
  readln(ex,ey,n1);
  readln(ax,ay,bx,by);
  fac[]:=; fac[]:=; exf[]:=; exf[]:=;
  for i:= to  do fac[i]:=fac[i-]*i mod mo;
  for i:= to  do exf[i]:=exf[mo mod i]*(mo-mo div i) mod mo;
  for i:= to  do exf[i]:=exf[i-]*exf[i] mod mo;

  eps:=1e-8;
  v:=(ex*ay-ey*ax)/(bx*ay-by*ax);
  u:=(ex*by-ey*bx)/(ax*by-ay*bx);
  if (fabs(round(v)-v)>eps)or(fabs(round(u)-u)>eps) then
  begin
   writeln();
   close(input);
   close(output);
   exit;
  end
   else begin inc(n); x[n]:=round(u); y[n]:=round(v); end;
  for i:= to n1 do
  begin
   readln(sx,sy);
   v:=(sx*ay-sy*ax)/(bx*ay-by*ax);
   u:=(sx*by-sy*bx)/(ax*by-ay*bx);
   if (fabs(round(v)-v)>eps)or(fabs(round(u)-u)>eps) then continue;
   if (u<)or(v<) then continue;
   if (u>x[])or(v>y[]) then continue;
   inc(n); x[n]:=round(u); y[n]:=round(v);
  end;
  inc(n); x[n]:=; y[n]:=;
  qsort(,n);
  dp[]:=;
  for i:= to n do
  begin
   dp[i]:=c(x[i]+y[i],x[i]);
   for j:= to i- do
    if (x[i]>=x[j])and(y[i]>=y[j]) then
     dp[i]:=dp[i]-dp[j]*c(x[i]+y[i]-x[j]-y[j],x[i]-x[j]) mod mo;
   dp[i]:=(dp[i] mod mo+mo) mod mo;
  end;
  writeln(dp[n]);

  close(input);
  close(output);
 end.