一道纯的高精度题，就是靠基础代码了，1AC0ms无压力。
注意整数和小数的分界，还有0的时候。
View Code 1 program pku1001(input,output); 2 type 3 number = array[0..100] of integer; 4 var 5 power : integer; 6 s,s1,s2 : ansistring; 7 x,answer : number; 8 n : longint; 9 procedure init; 10 begin 11 readln(s); 12 s1:=copy(s,1,pos(' ',s)-1); 13 delete(s,1,pos(' ',s)); 14 val(s,power); 15 n:=length(s1)-pos('.',s1); 16 delete(s1,pos('.',s1),1); 17 n:=n*power; 18 while s1[1]='0' do 19 delete(s1,1,1); 20 end; { init } 21 procedure change(); 22 var 23 i : longint; 24 begin 25 x[0]:=length(s1); 26 for i:=1 to length(s1) do 27 x[i]:=ord(s1[length(s1)+1-i])-48; 28 answer:=x; 29 end; { change } 30 function multiply(x,y :number ):number; 31 var 32 i,j : longint; 33 begin 34 fillchar(multiply,sizeof(multiply),0); 35 multiply[0]:=x[0]+y[0]-1; 36 for i:=1 to x[0] do 37 for j:=1 to y[0] do 38 begin 39 inc(multiply[i+j-1],x[i]*y[j]); 40 inc(multiply[i+j],multiply[i+j-1] div 10); 41 multiply[i+j-1]:=multiply[i+j-1] mod 10; 42 end; 43 if multiply[multiply[0]+1]>0 then 44 inc(multiply[0]); 45 end; { multiply } 46 procedure main; 47 var 48 i : longint; 49 begin 50 for i:=1 to power-1 do 51 answer:=multiply(answer,x); 52 s1:=''; 53 s2:=''; 54 for i:=answer[0] downto n+1 do 55 s1:=s1+chr(answer[i]+48); 56 for i:=n downto 1 do 57 s2:=s2+chr(answer[i]+48); 58 s:=s1+'.'+s2; 59 while s[1]='0' do 60 delete(s,1,1); 61 while s[length(s)]='0' do 62 delete(s,length(s),1); 63 if s[length(s)]='.' then 64 delete(s,length(s),1); 65 end; { main } 66 procedure print; 67 begin 68 if s='' then 69 writeln(0) 70 else 71 writeln(s); 72 end; { print } 73 begin 74 while not eof do 75 begin 76 init; 77 change; 78 main; 79 print; 80 end; 81 end.