Fzu_noip 1032 infinite number (carry determination)

Infinite Duration: 1 s memory: 32 M★Problem description: we generate two infinite numbers. The first number is a number that links all natural numbers. The second number is a number that connects the square of all natural numbers. Sum the two numbers as follows: 123456789101112131415161718192021... + 149162536496481100121144169196225... = 272619325597593231536305887388246... now I will give you an integer k. What is the number of k digits from left to right?★Data input: there are multiple groups of input data. Each group of data is input in one row, with a k number. For 100% of data, k <= 2147483647★Result output: for each group of data, an integer N is output, which is the k-digit number from left to right. Input sample output example 56781932 first calculates the k-bit Ai and Bi, then adds them, and determines whether to overhead them. Next, use search_fi (k: longint) to determine whether the k-th digit will make the previous carry. Then, [delphi] const a: array [1 .. 19, 1 .. 2] of int64 = (100,316), (317,999 ), ), (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;} 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. infinite Duration: 1 s memory: 32 M★Problem description: we generate two infinite numbers. The first number is a number that links all natural numbers. The second number is a number that connects the square of all natural numbers. Sum the two numbers as follows: 123456789101112131415161718192021... + 149162536496481100121144169196225... = 272619325597593231536305887388246... now I will give you an integer k. What is the number of k digits from left to right?★Data input: there are multiple groups of input data. Each group of data is input in one row, with a k number. For 100% of data, k <= 2147483647★Result output: for each group of data, an integer N is output, which is the k-digit number from left to right. Input sample output example 56781932 first calculates the k-bit Ai and Bi, then adds them, and determines whether to overhead them. Next, use search_fi (k: longint) to determine whether the k-th digit will make the previous carry. Then, [delphi] const a: array [1 .. 19, 1 .. 2] of int64 = (100,316), (317,999 ), ), (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;} 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.