optifine mod

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total Submission (s): 14494 Accepted Submission (s): 4484Problem descriptiongive a number n, find the minimum x (x>0) that satisfies 2^x mod n = 1.Inputone positive integer on each line, the value of N.Outputif the minimum x exists, print a line with 2^x mod n = 1.Print 2^? MoD n = 1

We assume that P is prime, n! =a*pe, we need to solve a mod p and E.E is n! The number of times that divisible p can be iterated, so use the following formula:N/p+n/p2+n/p3 ...We only need to calculate the pt≤n T, so the complexity is O (LOGPN)Next calculate a mod p.First calculate the n! The product of an item that cannot be divisible by P in the factor.As a simple example, it is not difficult to find that

Ubuntu seems to be directly supported, and now back to Debian a bit out of the way. Need to be manually configured:Installation module: Apt-get Install Libapache2-mod-pythonEdit file:/etc/apache2/sites-available/000-defaultAdd the following content: Restart Service:/etc/init.d/apache2 restartReference: Https://www.howtoforge.com/embedding-python-in-apache2-with-mod_python-debian-etchDebian Configuration Apache2 Support

DescriptionFermat ' s theorem states that for any prime number P and for any integer a > 1, ap = a (mod p). That's, if we raise a to the pth power and divide by P, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, the Some a. (and some, known as Carmichael Numbers, is base-a pseudoprimes for all a.)Given 2 P ≤1000000000 and 1 a p, determine whether or not p -is A base-a pseudoprime.InputInput contains se

In Oracle, the remainder is obtained using the MOD (numer1, number2) function, and the returned value is the other values. For example: Output 1 ~ All integer values in 13 that are not multiples of 3. Declare -- I number; begin for I in 1 .. 13 loop if Mod (I, 3 )! = 0 then dbms_output.put_line ('integer not a multiple of 3: '| + I); end if; end loop; Exception when others then dbms_output.put_line (sqlco

According to the algorithm of the payment card check digit in ISO 2894 the Luhn Mod-10 method stipulates: 1. Multiply each digit on the card number by the weight. The rule is that if the number of card number is even, then the first bit is multiplied by 2, otherwise it is multiplied by 1, then respectively, 1,2,1,2,1,2;2, if each digit multiplied by the weight after more than 9, you need to subtract 9;3. Sum all the weighted numbers processed, and us

Simulate the speed of the HTTP network when debugging the network using the Apache plug-in bandwidth mod 1. Install mod_bw Download http://bwmod.sourceforge.net /# Extract Enter the directory for execution Sudo apxs2-I-a-c mod_b1_c 2. Configuration Add the following to the configuration file/etc/apache2/httpd. conf: 1 kb/s in low speed mode Set to high-speed mode kb/s 3. The configuration takes effect. Sudo/etc/init. d/apache2 restart Reload doe

Originally postedIwantandroid from xda Starting with cyanogen mod for Nexus One This is only for system files, the instructions for Kernel making will come as soon as I get some time. Credits (these people helped me build form source the first time and created some awesome wiki at wiki.cyanogenmod.com)CyanogenMaxismaChris soyarsKmobsThedudesandroidRikupwPershootBcrookAnd more if I have missed just tell me If u want to learn building from source, c

Http://forum.androidspin.com/showthread.php/3973-ADVANCED-How-To-Mod-BOOT-amp-RECOVERY [Advanced] How to mod: boot Recovery This is heavily based on:Howto: unpack, edit, and re-pack boot images Android magic. IMG file is special (it is not regular, it has magic), it is the image file of a particle NAND partition... Use Linux GetMkbootfsMkbootimg These are tools made when compiling Android

Often we use Excel to participate in the operations between many data, such as: subtraction. Happens to be division, sometimes there will be no more than the number. What is the way to show this? The remainder is calculated directly? In fact, we can use the MoD function to do the operation. Calculates the remainder of the division of two numbers directly. Operation Steps Open Excel2010, and enter dividend and divisor separately in the A2, B2 cell bo

voidEXTENDGCD (Long LongALong LongBLong Longd,Long Longx,Long Longy) { if(b==0) {d=a;x=1; y=0;return;} EXTENDGCD (B,a%b,d,y,x); Y-= x* (A/b);}//solve A+c*x=b (mod D), return the smallest non-negative integer xLong LongModX (Long LongALong LongBLong LongCLong LongD) { if(a==B) {return 0; } if(c==0) { return-1; } Long Longx,y,tmpd; EXTENDGCD (C,d,tmpd,x,y); if((b-a)%tmpd! =0 ) { return-1; } Else{x*= (B-A)/tmpd; Long

Question: Returns a positive integer n, and returns the smallest x so that 2 ^ x mod n = 1. Ideas: N = 1. All positive mod 1 values are 0. N is an even number, why? The above formula can be deformed as follows: 2 ^ x = k * n + 1. If n is an even number, k * n + 1 is an odd number, and 2 ^ x is an even number. If n is an odd number, there must be a solution. For multiplication inverse element: in the op

1008 N's factorial mod PBase time limit: 1 seconds space limit: 131072 KB score: 0 Difficulty: Basic question input N and p (p is prime), N! Mod P =?(Mod is modulo%) For example: n = ten, P = 11,10! = 36288003628800% = 10InputTwo number n,p, separated by a space in the middle. (N OutputOutput n! The result of mod p.Inp

Fedora ( Linux platform) Open-source PDF file editing Tools-- PDF MoD650) this.width=650; " src="Http://timg01.bdimg.com/timg?pacompressimgtype =1sec=1439619614autorotate=1di=6aa2aa331c4b725018bb05f8a572affbquality=90size=b870_ 10000src=http%3a%2f%2fbos.nj.bpc.baidu.com%2fv1%2fmediaspot%2f9ab2f0cbb0df302473d3c93f89cfb8aa.jpeg " alt=" timg?pacompressimgtype=1sec=1439619614 " >650) this.width="650;"" > PDF files have very good cross-platform properties, regardless of which platform you open w

Description Given A,b,c, you should quickly calculate the result of A^b mod C. (1 Input There are multiply testcases. Each testcase, there are one line contains three integers a, B and C, separated by A. Output For each testcase, output a integer, denotes the result of A^b mod C. Sample Input 3 2 4 2 10 1000 Sample Output 1 24 train of Thought An exponential very scary topic, obviously fast power is no

uper a^b MoD CTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d %i64 U SubmitStatusDescriptionGiven A,b,c, you should quickly calculate the result of A^b mod C. (1InputThere is multiply testcases. Each testcase, there are one line contains three integers A, B and C, separated by A single space.OutputFor each testcase, output an integer, denotes the result of A^b

problem 1752 a^b mod Caccept:840 submit:3625Time limit:1000 mSec Memory limit:32768 KB problem DescriptionGiven A,b,c, you should quickly calculate the result of A^b mod C. (1InputThere is multiply testcases. Each testcase, there are one line contains three integers A, B and C, separated by A single space.OutputFor each testcase, output an integer, denotes the result of A^b

problem 1752 a^b mod Caccept:837 submit:3615Time limit:1000 mSec Memory limit:32768 KB problem DescriptionGiven A,b,c, you should quickly calculate the result of A^b mod C. (1InputThere is multiply testcases. Each testcase, there are one line contains three integers A, B and C, separated by A single space.OutputFor each testcase, output an integer, denotes the result of A^b

Difference between % and mod: the number of % comes from positive and negative, and the symbol depends on the left operand... MoD can only be positive (because a = B * q + R (q> 0 and 0 Therefore, if % is used to calculate mod, the formula is as follows:A mod B = (a % B + B) % B The goal in parentheses is to conve

question 1: Modlist not found This problem for most people in WIN10 can not solve. But after careful investigation by the landlord found The cause of the problem is ridiculous. Win10 automatically hides the suffix of the modlist. It's so good that the MoD doesn't recognize it. Question 1 Solution: Name the Modlist file as Modlist.txt.txt (note that two. txt) Perfect solution Question 2: Dotnet not installed or