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.