• BZOJ1089 [SCOI2003]严格n元树


    又是一道奇怪的DP(?)题
    一个非常好的想法是:
    令f[i]表示深度小于等于i的n元树的总个数,于是
    f[i] = f[i - 1] ^ n + 1 (这是因为加了一层以后新的根的n个儿子可以随便选,再加上没有儿子的情况)
    但是还要写高精。。。还好一边A了,手感不错~

      1 /**************************************************************
      2     Problem: 1089
      3     User: rausen
      4     Language: Pascal
      5     Result: Accepted
      6     Time:32 ms
      7     Memory:1204 kb
      8 ****************************************************************/
      9  
     10 const
     11   m : longint = 10000;
     12  
     13 type
     14   arr = array[0..10000] of longint;
     15  
     16 var
     17   f : array[0..20] of arr;
     18   ans : arr;
     19   i, n, d : longint;
     20  
     21 operator + (const a : arr; const b : longint) c : arr;
     22 var
     23   l : longint;
     24  
     25 begin
     26   fillchar(c, sizeof(c), 0);
     27   for l := 1 to a[0] do
     28     c[l] := a[l];
     29   c[1] := c[1] + b;
     30   l := 1;
     31   while c[l] >= m do begin
     32     c[l + 1] := c[l + 1] + c[l] div m;
     33     c[l] := c[l] mod m;
     34     inc(l);
     35   end;
     36   l := a[0] + 1;
     37   while c[l] = 0 do dec(l);
     38   c[0] := l;
     39 end;
     40  
     41 operator * (const a : arr; const b : arr) c : arr;
     42 var
     43   i, j, x, l : longint;
     44  
     45 begin
     46   fillchar(c, sizeof(c), 0);
     47   for i := 1 to a[0] do
     48     for j := 1 to b[0] do begin
     49       x := i + j - 1;
     50       c[x] := c[x] + a[i] * b[j];
     51       c[x + 1] := c[x + 1] + c[x] div m;
     52       c[x] := c[x] mod m;
     53     end;
     54   l := 0;
     55   while l <= a[0] + b[0] + 1 do begin
     56     c[l + 1] := c[l + 1] + c[l] div m;
     57     c[l] := c[l] mod m;
     58     inc(l);
     59   end;
     60   l := a[0] + b[0] + 1;
     61   while c[l] = 0 do dec(l);
     62   c[0] := l;
     63 end;
     64  
     65 operator - (const a : arr; const b : arr) c :arr;
     66 var
     67   l, del : longint;
     68  
     69 begin
     70   fillchar(c, sizeof(c), 0);
     71   l := 1;
     72   del := 0;
     73   while l <= a[0] do begin
     74     c[l] := a[l] - b[l] - del;
     75     if c[l] < 0 then begin
     76       del := 1;
     77       c[l] := c[l] + m;
     78     end else del := 0;
     79     inc(l);
     80   end;
     81   l := a[0];
     82   while c[l] = 0 do
     83     dec(l);
     84   c[0] := l;
     85 end;
     86  
     87 procedure refresh(var a : arr);
     88 begin
     89   fillchar(a, sizeof(a), 0);
     90 end;
     91  
     92 function pow(a : arr; x : longint) : arr;
     93 var
     94   b : arr;
     95  
     96 begin
     97   refresh(b);
     98   b := b + 1;
     99   while x > 0 do begin
    100     if x and 1 = 1 then b := b * a;
    101     a := a * a;
    102     x := x shr 1;
    103   end;
    104   exit(b);
    105 end;
    106  
    107 procedure print(a : arr);
    108 var
    109   i : longint;
    110  
    111 begin
    112   for i := a[0] downto 1 do begin
    113     if i <> a[0] then begin
    114       if a[i] < 1000 then write(0);
    115       if a[i] < 100 then write(0);
    116       if a[i] < 10 then write(0);
    117     end;
    118     write(a[i]);
    119   end;
    120 end;
    121  
    122 begin
    123   readln(n, d);
    124   refresh(f[0]);
    125   f[0] := f[0] + 1;
    126   for i := 1 to d do
    127     f[i] := pow(f[i - 1], n) + 1;
    128   if d = 0 then writeln(1) else begin
    129     ans := f[d] - f[d - 1];
    130     print(ans);
    131   end;
    132 end.
    View Code
    By Xs酱~ 转载请说明 博客地址:http://www.cnblogs.com/rausen
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  • 原文地址:https://www.cnblogs.com/rausen/p/4029946.html
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