best number theory books

(intA[],intM[],intk) { intN[K];//This can be deleted intMM =1;//least common multiple intresult =0; for(inti =0; I ) {mm*=M[i]; } for(intj =0; J ){ intL, J; EXOJLD (mm/M[J],-M[j], L, J); N[J]= M[j] * j +1;//1N[J] = mm/m[j] * L;//2 "Note" 1 and 2 values should be equal. Result + = n[j]*A[j]; } return(Result% mm + mm)% mm;//fall between (0, MM), this is written to prevent result initial negative, in this case can not be negative may be direc

Hdu 4861 Couple doubi (number theory), hdudoubi Link to the question: hdu 4861 Couple doubi Two people play the game. There are k balls on the table. The value of the I-th ball is 1i + 2i + second + (p − 1) I % p. The two players take turns, if the value of DouBiNan is large, YES is output; otherwise, NO is output. Solution:First, DouBiNan is obtained first, so the maximum value of the remaining value must

Time Limit: 3000/1000 ms (Java/other) memory limit: 65535/32768 K (Java/other) total submission (s): 15 accepted submission (s): 5 Font: times New Roman | verdana | georgiafont size: Drawing → problem descriptiongiven two positive integers A and N, satisfaing gcd (A, n) = 1, please find the smallest positive integer x with a ^ x limit 1 (mod N ). inputfirst is an integer T, indicating the number of test cases. T Each of following T lines contains two

HDU 4983 goffi and GCD Idea: For the number topic, if K is 2 and N is 1, then only one type is possible. If other K> 2 is 0, you only need to consider k = 1, when k = 1, the factor n is enumerated, and then equal to the number of satisfied factors, then gcd (x, n) = The number of this factor is PHI (N/this factor), and then use the multiplication principle to cal

Question Link A student dreamed of walking in a street with many bars. He can have a drink at each bar. All bars have a positive integer number. This person can go from bar n to the bar where the number can be divided by N. Now he wants to go from bar a to bar B. How much wine can he have at most. Idea: Because B must be a multiple of A, which is multiplied by a from the beginning. In fact, it is to find th

Qingdao Race Training A-fantasy of a summation LightOJ-1213 time: October 5 23 o'clock 30Https://cn.vjudge.net/contest/258301#problemThe problem, test instructions gives is a K-cycle, each loop I weighs 1 to n, the loop body is res = (res + A[I1] + A[i2] + ... + a[ik])% MOD; The value of the last res.Analysis: Combinatorial math + fast PowerSolution: For each a[i] is the same,So we figure out the total a[i] The number of occurrences/n to know the

LightOJ 1215 Finding LCM (number theory), lightojlcm It is known that LCM (a, B, c) = L and c. After expanding the number into the form of prime factor Product GCD (a, B) is a small index of the common prime factor of a and B. LCM (a, B) is a larger index of all prime factors of a and B. So that m = LCM (a, B) then the problem is converted to finding the minimum

The Euler function $phi (n) $ represents no more than the number of $n$ coprime in a positive integer of $n$ and has:$\varphi (n) = n\sum\limits_{p|n} (1-{\frac 1{p}}) $There are obviously $n$ primes:$\varphi (n) =n-1$and consider $mp$, if the $p$ is a prime number, then an arbitrary integer $k$:$ (MP, K) \leftrightarrow (M, K) $So there is $\varphi (m) $ number

Split Prime andTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d %i64 U DescriptionTo split an even number into two different prime numbers, how many methods are there?InputThe input contains some positive even numbers, the value will not exceed 10000, the number will not exceed 500, if in the event of 0, the end.Outputcorresponding to each even, the output is split into the

Tanabata FestivalTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d %i64 U DescriptionOn the day of Tanabata, matchmaker came to the digital kingdom, and he put a sign on the city gate and said to the people of the digital kingdom, "Do you want to know who your other half is?" then follow the signs and find it!People come to the notice before they want to know who is their other half. The notice is as follows: The factor of the number