題意:對于一個16進制數x,把x的各個數位拿出來,設其為t1,t2,...,定義s(x)為2^t1|2^t2|...,如x=0x3e53,則s(x)=2^3|2^14|2^5|2^3=16424.給出q組詢問l,r(l,r也是16進制數,不超過15位),求[l,r]中有多少個數x滿足x^s(x)<x.

這題題解寫的是個狀壓數位dp,但是蒟蒻不會數位dp,自己YY了一個做法.

首先,將[l,r]的詢問拆成2個前綴和,考慮到十六進制數減1并不方便,可以轉化為solve(r)-solve(l),再特判一下l滿不滿足要求.

考慮求出[0,upper]間的答案.將二進制位從低位向高位,從0開始編號.觀察x和s(x)的二進制表示,可以發現當且僅當s(x)的最高位1所在的數位上x也為1時,x^s(x)<x.考慮從0到15枚舉這個數位的位置id,下面只需求出不超過upper的數中有多少個數在id位為1,且所有數位都不超過id,且有至少一個數位等于id.發現最后一個條件不好處理,可以將詢問拆成2個:設solve(upper,id,lim)表示不超過upper的數中有多少個數在id位為1,且所有數位都不超過lim.所求答案即為solve(upper,id,id)-solve(upper,id,id-1).這個東西是可以掃幾遍upper討論一通求出來的.實現上細節非常惡心比較多.

 1 program cf776G;
 2 const number=['a'..'f','0'..'9'];
 3       list:array[0..15]of char=('0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f');
 4 var q,i:longint;
 5     l,r:string;
 6     num:array[char] of longint;
 7     ch:char;
 8 procedure getstr(var s:string);
 9 var ch:char;
10 begin
11   s:='';
12   read(ch);
13   while not (ch in number) do read(ch);
14   while ch in number do
15   begin
16     s:=s+ch;
17     read(ch);
18   end;
19 end;
20 function check(const x:string):longint;
21 var i:longint;
22     t,this:int64;
23 begin
24   this:=0;
25   for i:=1 to length(x) do this:=this*16+num[x[i]];
26   t:=0;
27   for i:=1 to length(x) do t:=t or (1 shl num[x[i]]);
28   exit(ord(this xor t<this));
29 end;
30 function solve2(const upper:string;id,up:longint):int64;
31 var upp,ans:int64;
32     i,this,t:longint;
33     sw:array[0..30] of longint;
34     pow:array[0..20] of int64;
35 begin
36   upp:=0;
37   for i:=1 to length(upper) do upp:=upp*16+num[upper[i]];
38   for i:=1 to length(upper) do sw[length(upper)-i]:=num[upper[i]];
39   if(1 shl (id mod 4)>up)or(1 shl id>upp)or(up<0) then exit(0);
40   t:=0;
41   for this:=0 to up do if this and (1 shl (id mod 4))>0 then inc(t);
42   ans:=0;
43   pow[0]:=1;
44   for i:=1 to length(upper) do pow[i]:=pow[i-1]*(up+1);
45   for i:=length(upper)-1 downto 0 do
46     if i=id div 4 then
47     begin
48       if up<sw[i] then
49       begin
50         ans:=ans+t*pow[i];
51         exit(ans);
52       end;
53       for this:=0 to sw[i]-1 do if this and (1 shl (id mod 4))>0 then
54         ans:=ans+pow[i];
55       if sw[i] and (1 shl (id mod 4))=0 then exit(ans);
56     end else
57     begin
58       if up<sw[i] then
59       begin
60         if i>id div 4 then ans:=ans+pow[i]*t else ans:=ans+pow[i+1];
61         exit(ans);
62       end else
63       begin
64         if i>id div 4 then ans:=ans+pow[i-1]*sw[i]*t
65           else ans:=ans+pow[i]*sw[i];
66       end;
67     end;
68   inc(ans);
69   exit(ans);
70 end;
71 function solve(const upper:string):int64;
72 var ans:int64;
73     k:longint;
74 begin
75   ans:=0;
76   for k:=0 to 15 do
77     ans:=ans+solve2(upper,k,k)-solve2(upper,k,k-1);
78   exit(ans);
79 end;
80 begin
81   for i:=0 to 15 do num[list[i]]:=i;
82   readln(q);
83   for i:=1 to q do
84   begin
85     getstr(l);getstr(r);
86     writeln(solve(r)-solve(l)+check(l));
87   end;
88 end.

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轉載于:https://www.cnblogs.com/lkmcfj/p/6473927.html

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