number theory book

Portal Previous questions During the training, the teacher told jokes together. This question first look for a regular table on the guess is P-1) * (q-1)/4 Without a set of Sigma, the formula is first divided into two (in fact, the same) Then consider each item before the subsummation. For each K, the result is an integer in Y = Q/P * X. Then the other side is the complementary side. (The two parts of this formula are the integral points of the two parts divided by the main diagonal line of a sq

Question: hdu4944fsf's game (number theory) Given N, there will be n * (n + 1)/two class rectangles (1*1, 2*1, 2*2 ,... N * 1, N * 2 .. N * n ). divide these matrices into K * k squares of the same size, and you can obtain the gold coins a * B/gcd (/K, B/K); then, give n and ask about the total gold coins that can be obtained. Solution: sum (n): N * (n + 1)/sum (n-1) + f (N) :( N * 1, N * 2 ,... N * n) th

http://acm.tzc.edu.cn/acmhome/problemdetail.do?method=showdetailid=3012Test instructions:f[1]=1,f[2]=2,f[n]=f[n-1]+f[n-2], definition:, BegThis problem is only 1s, it is obvious that the s[x in the exponential part of the model is timed out, let alone exponentiation.and the f[X] is the Fibonacci sequence, so the idea is to construct a matrix that uses matrices to f[X] and s[x].The initial construction method is a 4*4 matrix:Recorded as a[n]=t * b[n-1], then you can get a[n]=t^ (x-2) * b[2] (when

HDU1098 Ignatius's puzzle [number theory], hdu1098ignatius Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission (s): 6559 Accepted Submission (s): 4540 Problem DescriptionIgnatius is poor at math, he falls into ss a puzzle problem, so he has no choice but to appeal to Eddy. this problem describes that: f (x) = 5 * x ^ 13 + 13 * x ^ 5 + k * a *

For a pair of numbers (p, q), if their GCD is x0 and the LCM is y0, P * q/X0 = y0, that is, q = x0 * Y0/P, Since p and q are positive integers, p and q must all be x0 * y0 divisor. Therefore, the O (SQRT (x0 * y0) enumerated approx. It is determined by GCD in sequence. 1 # include [Number Theory] [maximum public approx.] [enumeration approx.] codevs 1012 maximum public approx. and minimum public I i

Small M factor and time limit: 1000 MS | memory limit: 65535 kb difficulty: 2 Description When he was in class, Mr. M was a bit out of his mind, and the teacher was hard on his question. I heard that Mr. M is still very proud of finding the factor and the power of B. But after reading the question, I want to find the factor and the power of B. I am a little confused. Can you solve this problem? Input Multiple groups of test examples Each row has two numbe

Xdu 1022 (number theory screening), xdu1022 1022: A simple math problem 2 time limit: 1 Sec memory limit: 128 MB Submitted: 73 solution: 13 [Submit] [Status] [discussion version] Question description Gaussian function: [x] indicates the largest integer less than or equal to x, that is, the rounded down. For example, [2.5] = 2, [1.2] = 1. Define function f (n) = [n/1] + [n/2] + [n/3] +... + [n/n]. sum

x+p So we only need to calculate the smallest k for each B (0 So for each B build a point, for each AI, from B-to (B+ai)%p a length AI-side model P and then build the diagram, goodpay attention to calculate the answer contribution there,/d[i] words have honey juice re#include #include#include#include#includeusing namespaceStd;typedefLong Longll;Constll n=5*1e5+5, inf=1e19;ll n;ll P=inf,a[ -];; ll Bmx,bmn,ans=0;structedge{ll V,w,ne;} E[n* the];ll h[n],cnt=0;voidins (ll u,ll v,ll W)

2705: [Sdoi2012]longge problem time limit:3 Sec Memory limit:128 MBsubmit:1959 solved:1229[Submit] [Status] [Discuss] Descriptionlongge is very good at maths and he is very happy to challenge difficult math problems. Now the problem is: Given an integer N, you need to ask for ∑GCD (i, N) (16Sample Output15HINT"Data Range"For 60% of data, 0For 100% of data, 0SourceRound1 Day1ExercisesDirect Euler functions can ... (Note: Do not use a linear sieve, when you want to calculate Euler functio

Extended Euclidean algorithm.void exgcd (int a,int b,intx,inty) {if (!b) {X=1;y=0;return;} EXGCD (b,a%b,x,y); int temp=x;x=y;y=temp-a/b*y;}1) ax+by=c.The condition of the solution is C%GCD (A, b) ==0, because AX+BY=GCD (a, B) must have a solution.The solution is x0,y0, then the general solution x=x0+ (B/GCD (A, a) *t y=y0-(A/GCD (b)) *t2) ax≡1 (mod m)X is a on the inverse of M, i.e. Ax-my=1, if GCD (a,m)!=1 no solutionThe solution is x0, then the general solution x=x0+m*t A about m inverse is ab

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 explode ll so in the place of multiplication with

(int_=0,int__=0): First (_), second (__) {}BOOL operatorConstPep other)Const { returnFirst Other.first; }}namespacehash_table{#defineINF ~0u>>1#defineMAXN 10010structlinker{intHash,val; Linker*Next; Linker (int_,linker *__): Hash (_), Val (INF), next (__) {}}*FIR[MAXN]; Inlineinthash (intx) { intpos=x%MAXN; for(Linker *iter=fir[pos];iter;iter=iter->next)if(iter->hash==x)returnIter->Val; return(fir[pos]=NewLinker (X,fir[pos])Val; }}inline Pep EXGCD (intAintb) { if(!B)returnPEP

The array is too lazy to open, paste code it is simplevarN,k,m,p,x:int64;functionf (x,y:int64): Int64;begin ify=0 ThenExit1); F:=f (x,y>>1); F:=f*fMoDp; ifY and 1=1 ThenF:=f*xMoDp;End;beginread (n,k,m,p); X:= ((MMoDP) * (nMoDP)-(m* (m+1) >>1 MoDP) * (K-1))MoDp; ifx0 ThenX:=x+trunc (ABS (x)/p) *p+p; X:=xMoDp; Writeln (f (m,k-2) *xMoDp);End.No commercials, Just give me a second laugh. 233333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

Factstone BenchmarkTime limit:10000/5000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 1760 Accepted Submission (s): 973Problem Descriptionamtel have announced that it'll release a 128-bit computer chip by, a 256-bit computer by 2020, a nd so on, continuing its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in), a 32-bit computer in 1990, a 16-bit computer in 1980, an 8-bit computer in 1970 , and a 4-bit computer, its firs

Test instructions: least common multiple of n numberAnalysis: Using method: LCM (A,b,c) =LCM (A,LCM (b,c)). Note that the first addition to the multiplier prevents integer overflow (tips)Code:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.HDU 1019 Least Common multiple-number theory (least common multiple)

Divide and conquer the geometric series and die:1#include 2 using namespacestd;3 4typedefLong Longll;5 6 ll Pow_mod (ll A, ll B, ll MoD)7 {8ll ans =1, W = a%MoD;9 while(b)Ten { One if(B 1 ) A { -Ans = ans * w%MoD; - } theb = b >>1; -w = w * W%MoD; - } - returnans; + } - + ll sum (ll p, ll Q, ll MoD) A { at if(q = =0 ) - return 1; - if(Q 1 ) - returnSUM (p, Q/2, MoD) * (1+ Pow_mod (p, Q/2+1, MoD))%MoD; - Else - return(

---restore content starts--- T has a version .... Because I'm looking for factorization in the wrong position ...Consider the contribution of each factor of N to the answer. The answer is that ∑d * PHI (N/D) (d | n) directly enumerates the factors of N and then asks for Phi.But we can do better.Notice that h (n) = ∑d * PHI (N/D) (d | n) is the form of Dirichlet convolution, and f (x) = X and f (x) = Phi (x) are integrable functions, so the answer H (x) is also an integrable function.so h (x) = Π

#include #include #define LL __int64ll Mult_mod (ll A,ll b,ll c){A%=c;B%=c;LL ret=0;while (b){if (b1) {ret+=a;ret%=c;}aif (a>=c) a%=c;b>>=1;}return ret;}ll Powermod (ll A, ll B, ll c){LL ans = 1;LL k = a% c;while (b>0)//(k*k% c) 2^b%c{if (b1)Ans = mult_mod (ans,k,c);b = b>>1;K = Mult_mod (k,k,c);}return ans;}ll GCD (ll A,ll b){if (a==0) {return B;}return gcd (B%a,a);}int main (){int T;scanf ("%d", t);LL a,m,n,k,g;while (t--){scanf ("%i64d%i64d%i64d%i64d", a,m,n,k);if (m>n){G = gcd (n,m);}Else{G

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

A score if 3/5=1/(1+2/3) =1/(1+1/(1+1/2));When the molecule appears 1, just let the denominator subtract one.#include #include#include#defineEX (A, b) (a = a ^ B, b = a ^ B, a = a ^ b)using namespacestd;intgcdintAintb) { return(b = =0) ? A:GCD (b, a%b);}classfenshu{ Public: intx, y; Fenshu friendoperator+(Fenshu N, Fenshu m) {m.x*=N.Y; N.x*=m.y; M.y*=N.Y; N.Y=m.y; N.x+=m.x; intK =gcd (n.x, N.Y); N.x/=K; N.Y/=K; returnN; } Fenshu friendoperator/ (intK, Fenshu N) {EX (n.x, N.Y); N.x*=