vijos-p1223 Mason Number (high-precision multiplication + split + python)

P1223Mason Number accepted Tags: NOIP universal group 2003[Display label]<textarea id="code" class="textbox" style=""></textarea>Describe

The prime number of a shape such as 2^p-1 is called Mason, at which point P must also be a prime. But the reverse is not necessarily, that is, if P is a prime number, 2^p-1 is not necessarily prime. 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, calculate the number of bits of 2^p-1 and the last 500 digits (expressed in decimal high precision number)

Format input Format

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

Output format

First line: Decimal high-precision number of digits 2^p-1.

第2-11: The last 500 digits of the decimal high-precision number 2^p-1. (output 50 bits per line, output 10 lines, less than 500 bits when the high 0)
You do not have to verify that 2^p-1 and P are prime numbers.

Example 1 sample input 1[copy]

1279

Sample output 1[Copy]

3860000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001040 7932194664399081925240327364085538615262247266704805319112350403608059673360298012239441732324184842421613954281007791383 5662483234649081399066056773207629241295093892203457731833496615835504729594205476898112116936771475484788669625013844382 6029173234888531116082853841658502825560466622483189091880184706822220314052102669843548873295802887805086973618690071472 0710555703168729087

Limit

Each test point 1s

Tips

It's very simple, don't think it's complicated! ^_^

This topic, to use just one point, high-precision multiplication, if it is C + +, please use the method of partition, and for Java and Python, only need to directly call the function of high-precision calculation, here with the Python POW

The calculation of the number of bits is simple, 10^x + k = 2^p-1, log10 (2^p-k-1) = = x-> int (log10 (2) * p) + 1

Python code:

#!/usr/bin/env python3#-*-coding:utf-8-*-import mathimport sysp = Int (raw_input ()) print int (MATH.LOG10 (2) * P) + 1L =  Pow (2,p)-1L = l% pow (10,500) #将数字减少, otherwise the time of the remainder operation will increase, resulting in a timeout f = []for i in range:    f.append (L)    l/= 10for I In range (500-1,-1,-1):    sys.stdout.write ('%d '% f[i])    if i% = = 0:        print '        

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

vijos-p1223 Mason Number (high-precision multiplication + split + python)