fzu_noip 1032 (无穷数-进位判定)

无穷数

时限:1s内存:32M

★问题描述:

我们生成两个无穷大的数,第一个数是把所有的自然数链接起来组成的数字;第二个数是把所有自然数的平方连接起来组成的数。对这两个数求和,如下:

 123456789101112131415161718192021…

+ 149162536496481100121144169196225…

= 272619325597593231536305887388246…

现在给你一个整数k,问和从左往右数第k位的数码是多少?

★数据输入:

输入数据有多组,每组数据输入一行,有一个数k。对于100%的数据,k<=2147483647

★结果输出:

对于每组数据,输出一个整数N,从左往右数第k位的数码。

输入示例

输出示例

5

6

7

8

1

9

3

2

 
先算出第k位的Ai和Bi,然后相加,考虑是否加过头。
接下来用search_fi(k:longint) 判断第k位是否会令前一位进位,则


const
   a:array[1..19,1..2] of int64=((1,3),(4,9),(10,31),(32,99),(100,316),(317,999),(1000,3162),(3163,9999),(10000,31622),(31623,99999),(100000,316227),(316228,999999),(1000000,3162277),(3162278,9999999),(10000000,31622776),(31622777,99999999),
   (100000000,316227766),(316227767,999999999),(1000000000,2147483647));
var
   k:longint;
function search_ai(k:longint):longint;
var
   i,j,d,k2,g:int64;
begin
   d:=1;i:=9;
   while (true) do
   begin
      if (k-d*i>0) then
      begin
         dec(k,d*i);
         i:=i*10;inc(d);
      end
      else break;
   end;
   k2:=i div 9+(k-1) div d;
   g:=(k-1) mod d+1;
   g:=d-g+1;
   while (g>1) do begin k2:=k2 div 10;dec(g); end;
   exit(k2 mod 10);
end;
{
function search_bi(k:longint):longint;
var
   i,j:int64;
   head:longint;
begin
   head:=1;j:=10; i:=1;
   while (i<=2147483647 ) do
   begin
      if ((i*i) div j>0) then
      begin
         write('(',head,',',i-1,')',',');
         head:=i; j:=j*10;
      end;
      inc(i);
   end;
end;      }

function search_bi(k:longint):longint;
var
   i:longint;
   g,k2:int64;
begin
   i:=1;
   while (k>i*(a[i,2]-a[i,1]+1)) do
   begin
      dec(k,i*(a[i,2]-a[i,1]+1));
      inc(i);
   end;

   k2:=(k-1) div i+1;
   k2:=a[i,1]-1+k2;
   k2:=k2*k2;
   g:=(k-1) mod i+1;
   g:=i-g+1;

   while (g>1) do begin k2:=k2 div 10;dec(g); end;
   exit(k2 mod 10);
end;
function search_fi(k:longint):longint;
var
   i:longint;
begin
   i:=search_ai(k)+search_bi(k);
   if (i<=8) then exit(0)
   else if (i>9) then exit(1)
   else exit(search_fi(k+1));

end;
begin
//   assign(output,'dabiao.out');
//   rewrite(output);

   while not eof do
   begin
      readln(k);
      writeln((search_bi(k)+search_ai(k)+search_fi(k+1)) mod 10);

   end;

//   search_bi(1);
//close(output);
end.