JZOJ2017.08.15 C组

T1

题目描述

思路：

较复杂的模拟。详见代码。

代码：

var n,m,i,p,ans,j,l,r,c:longint;    a,b,t:array[0..10110] of char;begin  assign(input,'eat.in');  assign(output,'eat.out');  reset(input);  rewrite(output);  readln(n);  for i:=1 to n do begin read(a[i]); t[i]:=a[i]; end;  readln;  for i:=1 to n do read(b[i]); readln;  readln(m);  for i:=1 to m do  begin    readln(l,r,c);    c:=c mod(r-l+1);    for j:=l to r do    begin      p:=(j+c-l+1)mod(r-l+1);      if p=0 then p:=r-l+1;      a[l+p-1]:=t[j];    end;    t:=a;  end;  for i:=1 to n do if b[i]=a[i] then inc(ans);  write(ans);  close(input);  close(output);end.

T2

题目描述

思路：

用链表统计高度，从高到低枚举高度。

代码：

var n,i,x,y,j:longint;    s,v,t,max,ans,maxh,maxc:longint;    a:array[0..501,0..100001]of integer;    b,c:array[0..100001]of longint;begin  assign(input,'house.in');  assign(output,'house.out');  reset(input);  rewrite(output);  read(n);  for i:=1 to n do  begin    read(x,y);    inc(a[y,x]); inc(b[x]); inc(c[x],y);    if x>maxh then maxh:=x;    if y>maxc then maxc:=y;    ans:=ans+y;  end;  for i:=1 to maxc do    for j:=1 to maxh do      a[i,j]:=a[i,j]+a[i,j-1];  for i:=maxh downto 1 do  begin    s:=b[i]-1; v:=c[i]; t:=maxc;    while (t>0) do    begin      if s>=a[t,i-1] then      begin        dec(s,a[t,i-1]);        v:=v+a[t,i-1]*t;      end else      begin        v:=v+s*t;        break;      end;      dec(t);    end;    if v>max then max:=v;  end;  write(ans-max);  close(input);  close(output);end.

T3

题目描述

思路：

二分时间。如果使用除草机的时间>mid 则不合法。

代码：

var a:array[-10..100011] of longint;    n,m,i:longint;    f:boolean;procedure qsort(l,r:longint);var i,j,mid:longint;begin  if l>=r then exit;  i:=l; j:=r; mid:=a[(i+j) div 2];  repeat    while a[i]<mid do inc(i);    while a[j]>mid do dec(j);    if i<=j then    begin      a[0]:=a[i]; a[i]:=a[j]; a[j]:=a[0];      //b[0]:=b[i]; b[i]:=b[j]; b[j]:=b[0];      inc(i); dec(j);    end;  until i>=j;  qsort(l,j);  qsort(i,r);end;function find(l,r:longint):longint;var mid,i,t:longint;    //f:boolean;begin  if l>r then exit(l);  mid:=(l+r) div 2;  f:=true; t:=0;  for i:=1 to n do    if a[i]>mid then    begin      t:=t+(a[i]-mid) div m;      if a[i] mod m>0 then inc(t);      if t>mid then begin f:=false; break; end;    end;  if f then exit(find(l,mid-1)) else exit(find(mid+1,r));end;begin  assign(input,'foreat.in');  assign(output,'foreat.out');  reset(input);  rewrite(output);  read(n,m);  for i:=1 to n do read(a[i]);  qsort(1,n);  write(find(0,a[n]));  close(input);  close(output);end.

T4

题目描述

思路：

其实我们只用枚举三个物品就行了，另一个自然而然得出。

var a:array[0..40000]of longint;    b,d:array[0..15000]of int64;    c:array[0..15000,1..4]of int64;    i,j,n,m,t:longint;begin  assign(input,'magic.in');  assign(output,'magic.out');  reset(input);  rewrite(output);  read(n,m);  for t:=1 to m do begin read(a[t]); inc(b[a[t]]); end;  for t:=1 to (n-1) div 9 do  begin    fillchar(d,sizeof(d),0);    for j:=n-t downto 8*t+1 do d[j]:=d[j+1]+b[j]*b[j+t];    for j:=1 to n-9*t-1 do    begin      c[j,1]:=c[j,1]+b[j+2*t]*d[j+8*t+1];      c[j+2*t,2]:=c[j+2*t,2]+b[j]*d[j+8*t+1];    end;    fillchar(d,sizeof(d),0);    for j:=1 to n-9*t-1 do d[j]:=d[j-1]+b[j]*b[j+2*t];    for j:=n-t downto 8*t+1 do    begin      c[j,3]:=c[j,3]+b[j+t]*d[j-8*t-1];      c[j+t,4]:=c[j+t,4]+b[j]*d[j-8*t-1];    end;  end;  for i:=1 to m do    writeln(c[a[i],1],' ',c[a[i],2],' ',c[a[i],3],' ',c[a[i],4]);  close(input);  close(output);end.