how to mod roblox

(ch>='0'ch'9') x=x*Ten+ch-'0', ch=GetChar (); - returnf*x; the } -ll N,K,C[MAXN][MAXN],SM[MAXN][MAXN];intT; - voidCalc () { -c[0][0]=sm[0][0]=1; Inc (I,1, MoD) sm[0][i]=1; +Inc (I,1, MoD) { -d[n][0]=sm[i][0]=1; +Inc (J,1, i) c[i][j]= (c[i-1][j-1]+c[i-1][J])%MoD; AInc (J,1, MoD) sm[i][j]= (sm[i][j-1]+C[I][J])%

of the method, see the next chapter of the blog.1228 Sequence summationSource: Hackerrank Base time limit: 3 seconds Space limit: 131072 KB score: 160 Difficulty: 6-level algorithm problem collection attention T (n) = N^k,s (n) = t (1) + t (2) + ... T (n). Given N and K, ask for S (n). For example k = 2,n = 5,s (n) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55. As a result, output s (n) Mod 1000000007 results. InputLine 1th: A number t that represents the numbe

1732 Fibonacci Series 2time limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingTitle DescriptionDescriptionIn the "1250 Fibonacci series", we find the value of the nth Fibonacci series. But in 1250, nEnter a descriptionInput DescriptionEnter multiple sets of data, one row for each group of data, and an integer n (1 Output descriptionOutput DescriptionOutputs several rows. Output per line (corresponding input) n Fibonacci number (considering the number will be very large,

the Fp is a+b≡c (mod p); that is, the remainder of (a+c) ÷p is the same as the remainder of C÷p.The rule of the Fp multiplication (AXB) is axb≡c (mod p);The Fp Division (a÷b) rule is a/b≡c (mod p), which is axb-1≡c (mod p); (B-1 is also an integer between 0 and p-1, but satisfies the bxb-1≡1 (

", "latest article" These contents, all when he is MC1,MC2,MC3, the experiment begins with the simplification first! Open Flash, do a similar figure in the MC,MC has a dynamic text box, variable name: Mctext, at the same time to the MC plus the corresponding link identification ======================flash Source Code ======================//Set margin First, 10px var border:number = new number (10); var xmodel = border; var mod:array = new Array (); This.attachmovie ("MC", "MC", This.getnexthig

The remainder of 100 divided by 7 is 2, meaning that 100 items seven seven are divided into one group and the last 2 are left. The remainder has a strict definition: if dividend is a and the divisor is B (assuming they are all positive integers), then we can always find a natural number less than B r and an integer m, making the a=bm+r. This r is the remainder of a divided by B, and M is called quotient. We often use MoD to represent the remainder, a

sides of the equation by d to get a/d * x + B/d * y = c/d. Obviously, a is the division of d, and B is the division of d, x and y are Integer Solutions, So c/d is also an integer. If c does not divide d, of course it is Impossible. Otherwise, if we can find the solutions x0 and y0 for ax0 + by0 = d, multiply the two sides by c/d, that is, a (c/d * x0) + B (c/d * y0) = c, you can get the solution of the original equation x = (c/d * x0), y = (c/d * y0. Theorem involved: ax+by=c（1）There is a group

, $ Z(Store OCTA): S (M8 [a]) ⟵ S ($ X ). A2Stbu $ X, $ y, $ Z(Store byte unsigned): U (m1 [a]) ⟵ U ($ X) mod 28. A6Stwu $ X, $ y, $ Z(Store wyde unsigned): U (m2 [a]) ⟵ U ($ X) mod 216. AASTTU $ X, $ y, $ Z(Store Tetra unsigned): U (M4 [a]) ⟵ U ($ X) mod 232. AESTOU $ X, $ y, $ Z(Store Octa unsigned): U (M8 [a]) ⟵ U ($ X ). B2Stht $ X, $ y, $ Z(Store high t

, secret key generation and encryption and decryption process1. Secret key generationEach user will generate their own public and private keys, with the following process:1) Select two large primes $p$ and $q$.2) Calculate the product $n=p \times q$ of $p$ and $q$.3) Randomly select a number coprime with $\phi (n) = (p-1) \times (q-1) $ $e$, which is $GCD (d, (p-1) \times (1-1)) =1$, which is usually selected in the application.4) Calculate $e$ modulo $\phi (n) $ for the modulo inverse element $

"The main topic"If you use F[i][j] to represent the elements of column I, J, in the Matrix, then F[i][j] satisfies the following recursion:F[1][1]=1F[i,j]=a*f[i][j-1]+b (j!=1) ①F[i,1]=c*f[i-1][m]+d (i!=1) ②The a,b,c,d in a recursive style is a given constant. F[N][M].IdeasGrind for a morning, however the extra data on the UOJ is not over yet. Bzoj on the AC first put up, follow slowly grind ...* There is a point, the final output of the answer when the first +

Topic linksThe value of C (n, m)%p, N, mFirst, the value of the C (n, m)%PI is calculated, and then this is an equation of congruence. Solved by the Chinese remainder theorem.#include #include#include#include#includeusing namespacestd;#definell Long Longll a[ -], b[ -];voidExtend_euclid (ll A, ll B, ll x, LL y) { if(b = =0) {x=1; Y=0; return; } extend_euclid (b, a%b, x, y); LL TMP=x; X=y; Y= tmp-(A/b) *y;} LL Mul (ll A, ll N, ll MoD) {a= (a%

, and the modulus (the amount of "overflow") is 12. Now we move the hour hand counterclockwise from 4 to 2, and move clockwise 10 to get the same result. The arithmetic formula is 4-2 and 4+10, at a glance 4-2 is obviously not equal to 4+10, then why the two results on the clock is the same?That's "modulo" in mischief. When the result of the operation exceeds the counting range, the modulo operation is performed, and the 4+10=14>12,12 is 2=4+ (-2) after the modulo is obtained. At this point you

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+ module image (which contains init_module and Cleanup_ Module) + other parameters (such as Deps) of the collection body .3. Then, the system calls Create_module to create a module data structure in the kernel and "subscribe" to the required system (kernel) space.4, finally, through the system call Init_module, the user space to complete the one-way connection module image into the kernel space, and then call the module named init_module function (in the Linux character device driver article is

Server side and client's key system is not the same, called Asymmetric Key system RSA algorithm is based on modulo operation x mod n, in fact: [(a mod n) + (b mod n)] mod n = (a+b) mod n[(a mod n)-(b

,128mData size:For 30% data: nFor 70% data: nFor 100% data: n(data has been strengthened ^_^)Sample Description:Therefore the output should be 17, 2 (mod 38=2)Ideas:The main problem is not to break the first ride or add, I can be a stack of the same doctrine, the stack of line tree ... Forgive me for incompetence.The general solution is to turn multiplication into addition.(S+flag) *lazy=s*lazy+flag*lazy.Attention:Data size, not only to open a long lo

Q are available. Step 3 determines the decryption key D:D * e = 1 mod (p-1) * (q-1) can easily compute d based on E, p, and Q. Step 4 exposes integers r and E, but does not expose D. Step 5 encrypts the plaintext P (p is an integer less than R) to ciphertext C, calculated as C = p^e mod R. Step 6 decrypts ciphertext C to plaintext p, and the method is: p = c^d modulo R. However only according to R and

]: Name of the moduleExECutE (without the. mod extension)-M [TEXT]: Parameter to pass to the module. This can be pasSedMultipleTimeS withDiffErent parameter each time and they will all be sent to the module (I. e.-M Param1-m Param2, etc .)-D: Dump all known modules-N [NUM]: Use for non-default TCP port number-S: Enable SSL-G [NUM]: Give up after trying to connect for NUM seconds (default 3)-R [NUM]: Sleep NUM seconds between retry attempts (default 3)

First, I will attach the matrix67 explanation: Certificate ----------------------------------------------------------------------------------------------------------------------------------------------------- The work of Miller and Rabin has taken a revolutionary step in the Fermat test and established the legendary Miller-Rabin algorithm. The new test is based on the following theorem: If P is a prime number, X is a positive integer smaller than P, and x ^ 2

http://www.ifrog.cc/acm/problem/1049I usually look for the law in these maths problems.First the violence simulates the previous ones, and then finds (x, y) = (x, y-1) + (x-1, y) gets.But it's useless. Because to get (x, y-1) These, but also recursive processing, will be GG.Then find the rule is C (x + y, y)-C (x + y, y-1)Cannot nothing more yy. Try to match the relationship between X and Y.Generally must be related to these two numbers, and 2 * x these few relationships.#include #include#includ