vj p1032题解

原题叙述

此题就是考你的高精熟练度！

首先，要作出一个很简单的递推判断：

从个位开始找循环，后两位的循环节必然为后一位的整数倍，依此类推，找出后K位的循环节。

于是思路就很简明了。

有三点需要注意：

1.读入的时候只须记录后k位，运算过程中也只须记录后k位，所以此题最多是10进制下100位的高精。

2.记录结果的时候要注意，结果可能是高精，所以不光要设置一个不记录长度的伪*高精，还要记录一个真*高精，用于输出。

3.如果自乘超过10次，就找不到循环了....具体为何，说实话，我说不清楚，但是大家想想9的n次方，也许能领悟到那么些东西。

好，代码如下：

  1 type
  2         arra=record
  3                 da:array[0..210]of longint;
  4                 l:longint;
  5         end;
  6 var
  7         k:longint;n,r:arra;
  8 function min(a,b:longint):longint;
  9 begin
 10         if a<b then exit(a) else exit(b);
 11 end;
 12 procedure init;
 13 var
 14         s,s1,s2:string;
 15         t,q,c,b,i:longint;
 16 begin
 17         
 18         readln(s);
 19         fillchar(n,sizeof(n),0);
 20         t:=pos(' ',s);
 21         val(copy(s,t+1,length(s)-t),k);
 22         s1:=copy(s,1,t-1);
 23         if k>length(s1) then q:=length(s1) else q:=k;
 24         for i:=1 to q do
 25                 begin
 26                         n.da[i]:=ord(s1[length(s1)-i+1])-ord('0');
 27                 end;
 28         n.l:=k;
 29         fillchar(r,sizeof(r),0);
 30         for i:=1 to n.l do r.da[i]:=n.da[i];
 31         r.l:=k;
 32         while (r.da[r.l]=0)and(r.l>1) do dec(r.l);
 33 end;
 34 function check(a,r:arra;k:longint):boolean;
 35 var
 36         i:longint;
 37 begin
 38         for i:=1 to k do
 39                 begin
 40                         if a.da[i]<>r.da[i] then exit(false);
 41                 end;
 42         exit(true);
 43 end;
 44 procedure mult(a,b:arra;var r:arra);
 45 var
 46         i,j:longint;c:arra;
 47 begin
 48         fillchar(r,sizeof(r),0);
 49         fillchar(c,sizeof(c),0);
 50         for i:=1 to min(a.l,k) do
 51                 for j:=1 to min(b.l,k) do
 52                         begin
 53                                 c.da[i+j-1]:=a.da[i]*b.da[j]+c.da[i+j-1];
 54                                 if c.da[i+j-1]>=10 then
 55                                         begin
 56                                                 c.da[i+j]:=(c.da[i+j-1] div 10)+c.da[i+j];
 57                                                 c.da[i+j-1]:=c.da[i+j-1] mod 10;
 58                                         end;
 59                         end;
 60         for i:=1 to k do r.da[i]:=c.da[i];
 61         r.l:=k;
 62 end;
 63 procedure accmult(a:arra;b:longint;var c:arra);
 64 var
 65         i:longint;
 66 begin
 67         fillchar(c,sizeof(c),0);
 68         c.l:=a.l+2;
 69         for i:=1 to c.l do
 70                         begin
 71                                 c.da[i]:=a.da[i]*b+c.da[i];
 72                                 if c.da[i]>=10 then
 73                                         begin
 74                                                 c.da[i+1]:=(c.da[i] div 10)+c.da[i+1];
 75                                                 c.da[i]:=c.da[i] mod 10;
 76                                         end;
 77                         end;
 78         while (c.da[c.l]=0)and(c.l>1) do dec(c.l);
 79 end;
 80 procedure find;
 81 var
 82         i,j,times:longint;b:boolean;c,m,d,z:arra;
 83 begin
 84 
 85         fillchar(c,sizeof(c),0);
 86         m:=n;
 87         c.l:=k;
 88         fillchar(z,sizeof(z),0);
 89         z.l:=1;z.da[1]:=1;
 90         if (r.l=1)and(r.da[1]=0) then
 91                 begin
 92                         writeln(1);
 93                         halt;
 94                 end
 95         else
 96           for i:=1 to k do
 97                 begin
 98 
 99                         times:=1;
100                         repeat
101                                 c:=r;
102                                 mult(r,n,r);
103                                 mult(c,m,d);
104                                 inc(times);
105                                 b:=check(m,d,i);
106                         until (b)or(times>10);
107                         if b then
108                                 begin
109                                         accmult(z,times-1,z);
110                                         n:=c;
111                                         r:=c;
112                                         n.l:=k;
113                                 end
114                         else
115                                 begin
116                                         writeln(-1);
117                                         halt;
118                                 end;
119                 end;
120         for i:=z.l downto 1 do write(z.da[i]);
121 
122 end;
123 begin
124         init;
125         find;
126 end.