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.
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posted on 2014-10-16 23:02  Xs酱~  阅读(196)  评论(0)  编辑  收藏  举报