無窮數
時限：1s內存：32M
★問題描述:
我們生成兩個無窮大的數，第一個數是把所有的自然數鏈接起來組成的數字；第二個數是把所有自然數的平方連接起來組成的數。對這兩個數求和，如下：
123456789101112131415161718192021...
+ 149162536496481100121144169196225...
= 272619325597593231536305887388246...
現在給你一個整數k，問和從左往右數第k位的數碼是多少？
★數據輸入：
輸入數據有多組，每組數據輸入一行，有一個數k。對於100%的數據，k<=2147483647
★結果輸出:
對於每組數據，輸出一個整數N，從左往右數第k位的數碼。
輸入示例
輸出示例
5
6
7
8
1
9
3
2
先算出第k位的Ai和Bi，然後相加，考慮是否加過頭。
接下來用search_fi(k:longint) 判斷第k位是否會令前一位進位，則
[delphi]
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. 無窮數
時限：1s內存：32M
★問題描述:
我們生成兩個無窮大的數，第一個數是把所有的自然數鏈接起來組成的數字；第二個數是把所有自然數的平方連接起來組成的數。對這兩個數求和，如下：
123456789101112131415161718192021...
+ 149162536496481100121144169196225...
= 272619325597593231536305887388246...
現在給你一個整數k，問和從左往右數第k位的數碼是多少？
★數據輸入：
輸入數據有多組，每組數據輸入一行，有一個數k。對於100%的數據，k<=2147483647
★結果輸出:
對於每組數據，輸出一個整數N，從左往右數第k位的數碼。
輸入示例
輸出示例
5
6
7
8
1
9
3
2
先算出第k位的Ai和Bi，然後相加，考慮是否加過頭。
接下來用search_fi(k:longint) 判斷第k位是否會令前一位進位，則
[delphi]
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.
