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。

//Power of Ceyptography
const
        inf='1.txt';
        start=1000000001;
type
        data=array[0..200]of longint;
var
        p: data;
        look: data;
        n, k: longint;

operator =(a, b:data)z: boolean;
var
        i: longint;
begin
  if a[0]<>b[0] then exit(false);
  for i := 1 to a[0] do if a[i]<>b[i] then exit(false);
  exit(true);
end;

operator <(a, b: data)z: boolean;
var
        i: longint;
begin
  if a[0]>b[0] then exit(false)
    else if a[0]<b[0] then exit(true)
      else
        for i := a[0] downto 1 do
          if a[i]>b[i] then exit(false)
            else if a[i]<b[i] then exit(true);
  exit(false);
end;

operator *(a, b: data)z : data;
var
        i, j, g: longint;
begin
  fillchar(z, sizeof(z), 0);
  z[0] := a[0] + b[0] - 1;
  for i := 1 to a[0]+1 do begin
    g := 0;
    for j := 1 to b[0]+1 do begin
      g := g + a[i]*b[j] + z[i+j-1];
      z[i+j-1] := g mod 10;
      g := g div 10;
    end;
  end;
  while z[z[0]+1]<>0 do inc(z[0]);
end;

function pow(aa, b: longint): data;
var
        base, a: data;
begin
  fillchar(a, sizeof(a), 0);
  while aa>0 do begin
    inc(a[0]);
    a[a[0]] := aa mod 10;
    aa := aa div 10;
  end;
  base := a; fillchar(pow, sizeof(pow), 0);
  pow[0] := 1; pow[1] := 1;
  while b>0 do begin
    if 1 and b=1 then pow := pow * base;
    base := base *base;
    b := b >> 1;
  end;
end;

procedure init;
var
        c: char;
        i: longint;
        a: data;
begin
  fillchar(p, sizeof(p), 0);
  k := 0;
  read(n);
  read(c);
  while not eoln do begin
    read(c);
    inc(p[0]);
    a[p[0]] := ord(c) - 48;
  end;
  for i := 1 to p[0] do
    p[i] := a[p[0]-i+1];
end;

function wei(x: longint): longint;
begin
  wei := 0;
  while x>0 do begin
    inc(wei); x := x div 10;
  end;
end;

procedure main;
var
        left, right, mid: longint;
        tmp: data;
        key: longint;
begin
  key := wei(n);
  left := 1; //right := p[0] div w + 2;
  right := start;
  while {true}left<=right do begin
    mid := (left+right)>>1;
    if wei(mid)*key>p[0] then begin
      right := mid -1; continue;
    end;
    tmp := pow(mid, n);
    if tmp=p then begin writeln(mid); exit end
      else if tmp<p then left := mid+1
        else right := mid-1;
  end;
  writeln(right);
end;

begin
  assign(input,inf); reset(input);
  while not seekeof do begin
    init;
    main;
  end;
end.