POJ 2109
题目大意：给出n，p， 求k^n=p;数据保证k，n都在合理范围内，p是高精度了。
解：ym c，有pow函数，直接开方就行了..我还是写了一个高精度+二分，气愤的是他居然有无解的情况，而且很容易爆数组上限，要加一个数位的判断，嘛，写的就这样了，有点乱。
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1 //Power of Ceyptography 2 const 3 inf='1.txt'; 4 start=1000000001; 5 type 6 data=array[0..200]of longint; 7 var 8 p: data; 9 look: data; 10 n, k: longint; 11 12 operator =(a, b:data)z: boolean; 13 var 14 i: longint; 15 begin 16 if a[0]<>b[0] then exit(false); 17 for i := 1 to a[0] do if a[i]<>b[i] then exit(false); 18 exit(true); 19 end; 20 21 operator <(a, b: data)z: boolean; 22 var 23 i: longint; 24 begin 25 if a[0]>b[0] then exit(false) 26 else if a[0]<b[0] then exit(true) 27 else 28 for i := a[0] downto 1 do 29 if a[i]>b[i] then exit(false) 30 else if a[i]<b[i] then exit(true); 31 exit(false); 32 end; 33 34 operator *(a, b: data)z : data; 35 var 36 i, j, g: longint; 37 begin 38 fillchar(z, sizeof(z), 0); 39 z[0] := a[0] + b[0] - 1; 40 for i := 1 to a[0]+1 do begin 41 g := 0; 42 for j := 1 to b[0]+1 do begin 43 g := g + a[i]*b[j] + z[i+j-1]; 44 z[i+j-1] := g mod 10; 45 g := g div 10; 46 end; 47 end; 48 while z[z[0]+1]<>0 do inc(z[0]); 49 end; 50 51 function pow(aa, b: longint): data; 52 var 53 base, a: data; 54 begin 55 fillchar(a, sizeof(a), 0); 56 while aa>0 do begin 57 inc(a[0]); 58 a[a[0]] := aa mod 10; 59 aa := aa div 10; 60 end; 61 base := a; fillchar(pow, sizeof(pow), 0); 62 pow[0] := 1; pow[1] := 1; 63 while b>0 do begin 64 if 1 and b=1 then pow := pow * base; 65 base := base *base; 66 b := b >> 1; 67 end; 68 end; 69 70 procedure init; 71 var 72 c: char; 73 i: longint; 74 a: data; 75 begin 76 fillchar(p, sizeof(p), 0); 77 k := 0; 78 read(n); 79 read(c); 80 while not eoln do begin 81 read(c); 82 inc(p[0]); 83 a[p[0]] := ord(c) - 48; 84 end; 85 for i := 1 to p[0] do 86 p[i] := a[p[0]-i+1]; 87 end; 88 89 function wei(x: longint): longint; 90 begin 91 wei := 0; 92 while x>0 do begin 93 inc(wei); x := x div 10; 94 end; 95 end; 96 97 procedure main; 98 var 99 left, right, mid: longint; 100 tmp: data; 101 key: longint; 102 begin 103 key := wei(n); 104 left := 1; //right := p[0] div w + 2; 105 right := start; 106 while {true}left<=right do begin 107 mid := (left+right)>>1; 108 if wei(mid)*key>p[0] then begin 109 right := mid -1; continue; 110 end; 111 tmp := pow(mid, n); 112 if tmp=p then begin writeln(mid); exit end 113 else if tmp<p then left := mid+1 114 else right := mid-1; 115 end; 116 writeln(right); 117 end; 118 119 begin 120 assign(input,inf); reset(input); 121 while not seekeof do begin 122 init; 123 main; 124 end; 125 end.
