theory fast retailing

A^b Time Limit: 1000MS Memory Limit: 65535KB 64bit IO Format: DescriptionAsk for A's B-square, modulo mod (1InputMultiple sets of inputs, one row per set of data, 3 positive integers, respectively a,b,modOutputOne row for each set of data output, for the answerSample Input2 10 100000005 100 10 2 37Sample Output102400Template problem, mainly considering the 1e18 of the huge, ordinary fast power will expl

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=5768Main Topic :T set of data, the l~r satisfies: 1. is a multiple of 7, 2. The number of%pi!=ai for n primes.Topic Ideas:"Chinese remainder theorem" "Tolerant principle" "fast multiplication" "Number theory"　Because they are prime numbers, they are 22 of each other, satisfying the conditions of the Chinese remainder theorem.The 7 is added to the prime n

Today's a question, bare ntt, but I will not, so White sent 50 points.So he came to learn about NTT.The surface is very simple, do not bother to post, that is not the point I want to say.The emphasis is on NTT, also called the Fast number theory transformation.In many problems, we may encounter the polynomial multiplication problem in the mode meaning, when the traditional

]=1; for(; k; k >>=1) { if(k1) C =Multiply (c,a); A=Multiply (a,a); } returnC;}voidInitmat (Mat A) {a.mat[0][0]=Ten; a.mat[0][1]=-1; a.mat[1][0]=1; a.mat[1][1]=0;}intMain () {//freopen ("In.txt", "R", stdin); intT; scanf ("%d",t); while(t--) { intN; scanf ("%d",N); if(n==1) printf ("9\n"); Else if(n==2) printf ("97\n"); Else{Initmat (a); A=quickpower (a,n-2); intAns= (a.mat[0][0]*98+a.mat[0][1]*Ten-1)%1024x768;//we're asking for s[n]-1. while(ans0) ans+=10

Question: given n (nFind the root, take an indicator of each element in the S set, and then generate the generating function f (x)So the answer is (f (x)) ^n (mod x^ (m-1), mod 1004535809)On the NTT with a polynomial quick power to do it.#include Bzoj 3992 Sdoi2015 Sequence Statistics fast number theory transformation

HDU 2256 Problem of Precision number theory matrix fast power, hdu2256 You must obtain the integer (√ 2 + √ 3) of 2n and then mod 1024. (√ 2 + √ 3) 2n = (5 + 2 √ 6) n If the value is calculated directly, double is used to store the value. When n is large, the loss of precision will increase and the expected result cannot be obtained. We found that (5 + 2 √ 6) n + (5-2 √ 6) n is an integer (the even power o

) __builtin_popcount (x) +typedefLong LongLL; -typedefLong DoubleLD; $typedef unsignedLong LongULL; $typedef pairint,int>PII; -typedef pairstring,int>psi; -typedef pairint,int>PLL; thetypedef mapstring,int>MSI; -typedef vectorint>VI;Wuyitypedef vectorint>VL; thetypedef vectorVVL; -typedef vectorBOOL>vb; Wu - ConstLL mod = in; About LL N; $ - ll Mul (ll X, ll Q) { -LL ret =1; - while(q) { A if(Q 1) ret = (ret * x)%MoD; +Q >>=1; thex = (x * x)%MoD; - } $ returnret; the } the

The principle of rapid number theory change (NTT) is actually the same principle as the fast Fourier transform. For the Fermat prime of a shape such as m= c*2^n+1, it is assumed that its original root is G. So deceive g^ (m-1) ==1 and just (m-1) can divide 2^n. So the NTT transform can be performed in the modulo p field. The rotation factor is g^ ((m-1)/n). The other principles are the same as the FFT princ

Calculates a^n% B, where a, B and N are all 32-bit integers.Fast power is over. Fast power the first thing is to know(a*b)%c = ((a%c) *b)%c, so deduced.(a^n)%b = (((((((((a%b) *a)%b) *a) ...%b) *a)%b (n times)This can only solve the a^n beyond the computer count range, the complexity is still not lowered.What about ^O^,BIT-MANIPULATION!!!!!!Detailed details of self-Baidu ^_^ (using two points of thought)Code:classSolution {/** @param A, B, n:32bit int

Topic Links:Move instantlyproblem DescriptionThere is an infinitely large rectangle, initially when you are in the upper left corner (i.e. the first row of the first column), each time you can select a lower right lattice, and teleport past (such as the red lattice from which can be directly teleport to the blue lattice), the firstnN rows of M-m columns have several schemes, the answer to the 1000000007 modulo . InputMultiple sets of test data.Two integers n,m (2OutputAn integer representing t

BZOJ 3992 Sdoi2015 sequential statistics fast number theory transformation, bzoj3992 Given n (n Evaluate the original root, obtain the index for each element in the S set, and then generate the f (x) function) The answer is (f (x) ^ n (mod x ^ m-1), mod 1004535809) NTT just needs to use a polynomial to quickly create a power. #include