best number theory books

;Const intMAXN = 2e6+Ten;Const intShift =1e6;intF[MAXN];intMain () {intA, B, C, D; while(cin>>a>>b>>c>>d) {intI, J, k, ans =0; if(a0b0c0d0|| A>0b>0c>0d>0) {cout0Continue; } memset (F,0,sizeof(f)); for(i=1; i -; i++) for(j=1; j -; J + +) F[a*i*i+b*j*j+shift]++; for(i=1; i -; i++) for(j=1; j -; J + +) ans+=f[-c*i*i-d*j*j+Shift]; cout -Endl; } return 0;}/*Number theory, transformation for

Boolean[3005]; intPrime[] =New int[450]; inttop; BigInteger p[]=Newbiginteger[3005]; BooleanVis[] =New Boolean[3005]; //wagered sieve method for calculating primes voidIsPrime () { for(inti = 0; i i) isprime[i]=true; isprime[0] = isprime[1] =false;//Initialize for(inti = 2; I //Sieve Method { if(Isprime[i]) { for(intj = i*i; J //The upper bound is too large to explode int{Isprime[j]=false; }}} Top= 0; for(inti = 0; I i)if(Isprime[i]) prime[top++] =i; } BigInteger D

) - { - returnb==0? A:GCD (b, a%b); the } - - voidDfsintCur=1, LL Div=1ll,intfa=-1) - { + if(div>x) - { +coutsum; AExit0); at return; - } - if(Num[cur] | | G[cur].size () 1 cur!=1)//It's a leaf node . - { -X=min (x, num[cur]*div); -lcm=div*lcm/gcd (lcm,div); in return; - } to for(intI=0; I) + { - if(g[cur][i]!=FA) theDFS (G[cur][i], div* (LL) (G[cur].size ()-1* (cur!=1) ), cur); * } $ }Panax Notoginseng intMain () - { the //fr

decomposition 6 has a 2, a 3.72 after decomposition there are 3 2, 2 3.Then for the prime number 2, 6 of a 2 is the lower limit, and 72 of 3 2 is the upper limit. X, Y, z must have a number to decompose a prime 2, must have another number to decompose 3 2, and then the remaining number of the 2 must be divided between

1200 congruence equation2012 Noip National League Improvement Grouptime limit: 1 sspace limit: 128000 KBtitle level: Diamonds Diamond SolvingView Run ResultsTitle DescriptionDescriptionThe minimum positive integer solution for ax≡1 (mod b) of the x congruence equation is obtained.Enter a descriptionInput DescriptionEnter only one row, containing two positive integers a, b, separated by a space.Output descriptionOutput DescriptionThe output has only one row containing a positive integer x0, or a

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

1. PrefaceIn fact, calculus or something already finished ... Let's have a morning. Permutation combination, function limit, derivative, definite integral, indefinite integral and differential all three hours clear, and then go to speak the mother function and so on all kinds of things go, can't keep up, in the afternoon again examination. Fortunately, it's a little quiet today. Then there may not be enough time to speak in detail, after the Newton-Leibniz formula, roughly pull a pull into the m

Discrete Logging Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 4236 Accepted: 1948 DescriptionGiven a prime P, 2 BL = = N (mod P)InputRead several lines of input, each containing p,b,n separated by a space.OutputFor each line print the logarithm to a separate line. If There is several, print the smallest; If there is none, print "no solution".Sample Input5 2 15

Sumdiv Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 15745 Accepted: 3894 DescriptionConsider natural numbers A and B. Let S is the sum of all natural divisors of a^b. Determine s modulo 9901 (the rest of the division of S by 9901).InputThe only line contains the natural numbers A and B, (0 OutputThe only line of the output would contain S modulo 9901.Sample Input2 3Sample Output15Hint2^3 = 8.The natural divisors of 8 a

Ultraviolet A 11388 GCD lcm (number theory) Question: Check whether there is a, B so that lcm (a, B) = L, gcd (a, B) = g, there is no output-1, there is output A, B, and a is as small as possibleAnalysis: Forced violence is impossible. The data is huge. LlU is used. There are two ideas here.Idea 1: a * B = g * lIf a is a multiple of a = g, enumerateThen judge whether g * l can divide a, and finally judge wh

Dalao blog, at least very good-looking.Because I am the number theory is slag, but the examination is really the test, had to learn early, early master.General GCD of the maximum male factor1 int gcd (int x,int y)2{3 return0 ? X:GCD (y, x% y)4 }View CodeBinary optimization gcd1InlineintBSGCD (intXinty)2 {3 if(x = = y)returnx;4 if(x y;5 if(! (X 1))//x-even Y-even gcd (x, y) =2*gcd (X/2,Y/2)

the method given by the question, obtain f [N] Through preprocessing, process T [N] from the right to the left, and add M [T [N] to the answer. update M [(T [N] + R * f [N]) % P. Description Special processing is required when the prime number in the question is 2 or 5, because 10 can be divisible by 2 or 5, and the remainder cannot be obtained in the form of 10 power. In this case, you only need to keep counting the

(3,3) =13-3+1=1, only one time, and so on one count is 5 number of multiples C (5,1) =5c (5,2) =10C (5,3) =10c (5,4) =5c (5,5) =15-10+10-5+1=1 six numbers C (6,1) =6c (6,2) =15c (6,3) =20c (6,4) =15c (6,5) =6c (6,6) =16-15+20-15+6-1= 1 then because the number is not more than 10, you can use the idea of enumerating subsets to do this topic. So use DFS. Finally, there is a place to pay attention is in the D

See the definition of the ACM-ICPC series of number theory, the modulo operation is such a son.Given a positive integer p, any integer n, there must be an equation: n = kp + R, where K, R is an integer, and 0≤r 1printf"(7) MOD5 =%d\n",7%5);2printf"( -7) MOD5 =%d\n",(-7)%5);3printf"(7) MOD ( -5) =%d\n",7%(-5));4printf"( -7) MOD ( -5) =%d\n",(-7)%(-5));The result is:It's totally different, it's subverting my

last number you get is the number of numbers that are not coprime with X.Because one of the most basic rules of the repulsion principle is that--To calculate the size of several sets of merge sets, we first calculate the size of all the individual collections , then subtract all the two sets that intersect , and then add the parts that intersect all the three sets, minus all four sets intersect , and so on

#defineEPS 1e-9#defineAll (x) X.begin (), X.end ()#defineINS (x) Inserter (X,x.begin ())#definefor (i,j,k) for (int i=j;i#defineMAXN 1005#defineMAXM 40005#defineINF 0X3FFFFFFFusing namespaceStd;typedefLong LongLL; LL I,j,k,n,m,x,y,t,big,cas,num;BOOLFlag; LL Cur,ans;BOOLprim[2000005]; LL ver[2000005]; voidGetprim (ll size) {ll m=SQRT (size+0.5); memset (Prim,0,sizeof(prim));//can be emptied according to the situationnum=0;//put the found prime number i

) +f (2) +f (3) +1, that is, the number of answers to F (6) is the sum of the number of answers to all the natural numbers that can be taken before it (6 can be taken by three natural numbers), The last plus 1 means that the number 6 itself is also an answer;3. So, we can know F (n) =f (1) +f (2) +......f (Trunc (N/2)) +1;4. Therefore, F (n) is required, we simpl

remainder is the answer.Fermat theoremIf P is a prime number, if p is not divisible by a, then a^ (p-1) ≡1 (mod p), if p can divide a, then a^ (p-1) ≡0 (mod p). If P is a prime number and a,p coprime, then the remainder of a (p-1) divided by P is constant equal to 1.Proof:Since P is a prime number and (a,p) = 1, φ (p) =p-1. a^ (p-1) ≡1 (mod p) is available from

squares, if it will be changed state odd number of times, can be opened. P.S. The initial state of the bulb is out of the data to see ... There's going to be a burst int.1#include 2#include 3typedefLong LongLL;4 5 intMain () {6 LL N;7 while(SCANF ("%lld", n)! = EOF N) {8ll k = (ll) sqrt (n1.0);9 if(K*k = = N) puts ("Yes");Ten ElsePuts"No"); One } A return 0; -}View CodeMultiplying by Rotation Shift Division3 numbers per l

! ) )。Known? (n) To find the lawN?(1?1/ p 1 )?(1?1/ p 2 )....(1?1/ p n )(P is the mass factor of N), so m! , the molecule is m! , the denominator is the 1-m of all prime numbers. ( 1 1 / p multiply product The answer here is to ask, m! Eliminate, getN!/∏(1?1/ p i )mod 1000000007, first preprocessing those tables, each time to calculate can#include Recently brush problems feel good, light attention to the problem of the feeling is particularly busy, summary has no