2003 Mason Number

Title Description Description

The prime number of a shape such as 2p-1 is called Mason, at which point P must also be a prime. But not necessarily, that is, if p is a prime number, 2p-1 is not necessarily a prime number. By the end of 1998, 37 Mason had been found. The biggest one is p=3021377, which has 909,526 bits. Mason number has many important applications, which are closely related to the complete number.

Task: Enter P (1000<p<3100000) from the file to calculate the number of digits of 2p-1 and the last 500 digits (expressed in decimal high precision number)

Enter a description Input Description

Only one integer P (1000<p<3100000) is included in the file

Output description Output Description

First line: Decimal high precision number 2P-1 digits.

第2-11 line: Decimal High precision number 2P-1 the last 500 digits. (output 50 bits per line, output 10 lines, less than 500 bits when the high 0)

You do not have to verify that 2p-1 and p are prime numbers.

Sample input Sample Input

1279

Sample output Sample Output

386

00000000000000000000000000000000000000000000000000

00000000000000000000000000000000000000000000000000

00000000000000104079321946643990819252403273640855

38615262247266704805319112350403608059673360298012

23944173232418484242161395428100779138356624832346

49081399066056773207629241295093892203457731833496

61583550472959420547689811211693677147548478866962

50138443826029173234888531116082853841658502825560

46662248318909188018470682222031405210266984354887

32958028878050869736186900714720710555703168729087

Data range and Tips Data Size & Hint

Exercises

High precision + number theory + two points.

Use the formula: 2 (Y) =sqr (2 (Y Div 2)), if Y is an even number without tube, if y is odd, then multiply 2. Formula for the number of digits (2 (y)): Trunc (log10 (2) *y) +1 (log10 function in math Math library).

Uses math;

var i,p:longint;

A,c:array[0..2001]of Longint;

Procedure F (k:longint);

var i,j:longint;

Begin

If K=0 then exit;

F (k Div 2);

Fillchar (C,sizeof (c), 0);

For I:=1 to A[0] do

For J:=1 to A[0] do

Begin

Inc (C[i+j-1],c[i+j-2] Div 10+a[i]*a[j]);

C[I+J-2]:=C[I+J-2] MoD 10;

End

c[0]:=a[0]*2-1;

While c[c[0]]>9 do

Begin

C[c[0]+1]:=c[c[0]] Div 10;

C[C[0]]:=C[C[0]] MoD 10;

Inc (C[0]);

End

A:=c;

If c[0]>500 then a[0]:=500;

If k mod 2=0 then exit;

Fillchar (C,sizeof (c), 0);

For I:=1 to A[0] do

Begin

C[i]:=a[i]*2+c[i-1] Div 10;

C[I-1]:=C[I-1] MoD 10;

End

C[0]:=A[0];

While c[c[0]]>9 do

Begin

C[c[0]+1]:=c[c[0]] Div 10;

C[C[0]]:=C[C[0]] MoD 10;

Inc (C[0]);

End

A:=c;

If c[0]>500 then a[0]:=500;

End

Begin

READLN (P);

Writeln (Trunc (log10 (2) *p) +1);

A[0]:=1;

A[1]:=1;

f (P);

Dec (a[1]);

For i:=500 Downto 1 do

if (i mod 50=1) then Writeln (A[i])

else write (a[i]);

End.

2003 Mason Number