best number theory books

(intI=2; i*i) { while(n%i==0) {n/=i; res-= (res/i); while(n%i==0) n/=i; } } if(n>1) Res-= (res/N); returnRes;}/********** Euler function **************/intEuler (intN) { intres=N; for(intI=2; i*i) { while(n%i==0) {n/=i; res-= (res/i); while(n%i==0) n/=i; } } if(n>1) Res-= (res/N); returnRes;}/** * * * Prime sieve method + Euler hit table **********/ll E[n+5],p[n+5];BOOLvis[n+5];voidinit () {memset (E,0,sizeof(e)); Memset (P,0,sizeof(p)); memset (Vis,false,

[NOIP training] [Law + number theory] Application of Euler's function, noip Euler Problem 1 [Topic] Given Obtain multiple groups of data. Question Proof: Except for an even number, the numbers of mutual quality are paired. By. Therefore, for the sum of each pair, there are a total of pairs. Then Problem 2 [Topic] Write on the first circle, w

also possible, but tstructmod{Long Longa[4][4]; MoD () {memset (A,0,sizeof(a)); }};mod Mul (mod a,mod b)//matrix multiplication{mod C; for(intI=1; i3; i++) for(intj=1; j3; j + +) for(intk=1; k3; k++) C.a[i][j]= (C.a[i][j]+a.a[i][k]*b.a[k][j])%1000000007; returnC;}voidMakeintN) {mod a,c; c.a[1][1]=1; c.a[2][1]=1; c.a[3][1]=1; a.a[1][1]=0; a.a[1][2]=1; a.a[1][3]=0; a.a[2][1]=0; a.a[2][2]=0; a.a[2][3]=1; a.a[3][1]=1; a.a[3][2]=0; a.a[3][3]=1; N++; while(n>0)//Fast Power

11371-number Theory for Newbies Time limit:1.000 seconds Http://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8category=24page=show_problem problem=2366 Given any positive integer, if we permute it digits, the difference between the number we get and the Given number would a Lways is divisible by 9. For

A few days ago, inadvertently visited the science network, which introduced many number theory problems, which is my hobby, so I used to write a free time for the calculation of numbers of the unsigned large integer class. The basic structure of a classClass CUSuperInt { public: //构造及析构函数 CUSuperInt(); CUSuperInt(DWORD dwValue); CUSuperInt(char* pszVal); CUSuperInt(CUSuperInt x);　 virtual ~CUSuperInt(); pr

Looking for prime number pairsTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 8771 Accepted Submission (s): 4395Problem description Goldbach Guess we all know a little bit. We're not trying to prove that, but we want to take an even number out of a set of numbers that can be represented within a programming language to find two primes so that they are eq

;q>>r>>s;c=-C; if(p>q| | r>s| | (!a!bc)) {cout0Continue;} if(!a| |!b) { if(!a) a=q-p+1; Else if(c%a==0 (C/A) 1; ElseA=0; if(!b) b=s-r+1; Else if(c%b==0 (c/b) 1; Elseb=0; coutEndl; Continue; } exgcd (A,b,d,x,y); if(c%d!=0) {cout0Continue;} K=c/d;x*=k;y*=K; A=a/d;b=b/D; z[0]= (q-x) *1.0/b; z[1]= (p-x) *1.0/b; z[2]= (y-s) *1.0/A; z[3]= (Y-R) *1.0/A; Sort (z,z+4); A=floor (z[2]); B=ceil (z[1]); cout1Endl; } return 0;}2. Direct Enumeration#include #include#definell Long Longusing

This is an examination of the problem of palindrome number, asking you to output the K palindrome number. In the process of doing the problem, we can find the regularity of the distribution of palindrome number: one number: 9, two digits: 9, three digits: 90, four digits: 90, five digits: 900, six digits: 900 ...#inclu

The main idea:--I can't see it myself.A joke--so much to tell what is φ and μ---But not normal φ and μ, some variants--The newly defined φ (1) = 0, the newly defined μ only computes odd prime numbers, and the number with the 2 factor is zero for the μ value.We first find the first and second questions, that is, the μ value is not equal to 0 of the partBecause of the definition of μ, the μ value is not equal to 0 when and only if the

' bad P ', where P is the smallest factor in K. Otherwise, it should output "good". Cases should is separated by a line-break.Sample Input143 10143 20667 20667 302573 302573 400 0Sample OutputGoodbad 11GOODBAD 23GOODBAD 31SourceNordic 2005Title: Given two numbers k and l,k are multiplied by two primes, if the prime number is less than L, the output bad and that prime number, otherwise the output good.First

+ 34 + 55 = 143. The second test case, 120 + 120 + 220 + 320 = 3487832979, and 3487832979 = 3*1000000009 + 487832952, so the output is 487832952. Theme It is clear that the sum of the First N values of each number in the Fibonacci series is obtained from the power of M. Ideas At that time, I didn't dare to write down the value range of M until CF 255 (446 C) had a line segment tree similar to the idea. The official editorial gave an explanation, is

Question A is hard to understand. The number of quality factors that output the product of 10 numbers not greater than 100000 A large number is not allowed. The time is 2 S. X. First, we will learn the knowledge of a number theory: Positive number N can be decomposed into P1

Definition of prime number: In a natural number greater than 1, the number that cannot be divisible by other natural numbers except for 1 and the integer itself. 1 and 0 are neither prime nor composite. Prime numbers are two concepts relative to composite number, which constitute one of the most basic definitions in

"Idea":http://blog.csdn.net/iamskying/article/details/4738838Solution Ideas:Now to analyze a problem, assuming that the 10-bit number is a, and the single-digit is an integer of B as AB, it is deducedAb*ab = (a*10+b) * (a*10+b) = 100*a*a+10*2*a*b+b*bAccording to the formula available: root (ab*ab) = A*a+2*a*b+b*b = (a+b) * (A+B); [Formula I]The same can be proven: root (Ab*ab*ab) = (a+b) * (a+b) * (A+B); [Formula Two]It can be seen that the root of th

/=i; } A[i]=J; } intE=i; for(i=2; i){ if(!a[i])Continue; Doublet=i; S*=(1.0-1.0/t); } printf ("%d\n",(int) s); for(intI=1; i0; } return 0;}Read the puzzle, optimize the codeAC Code:#include #includeusing namespacestd;intMain () {Long LongN,ans; while(cin>>N) { if(!n) Break; Ans=N; if(n%2==0){ while(n%2==0) n/=2; Ans/=2; } for(Long LongI=3; i*i2){ if(n%i==0){ while(n%i==0) n/=i; Ans=ans/i* (I-1); } } if(n>

LinkDescriptionYou task was to find minimal natural number N, so that n! contains exactly Q zeroes on the trail in decimal notation. As you know n! = 1*2*...*n. For example, 5! = contains one zero on the trail.InputInput starts with an integer T (≤10000), denoting the number of test cases.Each case contains an integer Q (1≤q≤108) in a line.OutputFor each case, print the case

Enter an n (1Lacus theorem http://blog.csdn.net/acm_cxlove/article/details/7844973A, B is a non-negative integer and P is a prime number. AB is written in P-system: a=a[n]a[n-1]...a[0],b=b[n]b[n-1]...b[0].The combined number C (A, B) and C (A[n],b[n]) *c (a[n-1],b[n-1]) *...*c (a[0],b[0]) MODP the sameSo in this topic C (n, x) will n and X into 2 C (0,0) =1,c (0,1) =0,c (1,0) =1,c (a) =1So if C (n,x) is 1,

]] =true; - if(i% primes[j] = =0) { -Mu[i * Primes[j]] =0; - Break; in}Else { -Mu[i * Primes[j]] =-Mu[i]; to } + } - } the } * $ intMain ()Panax Notoginseng { - Moblus (); the intN; + while(SCANF ("%d", n)! =EOF) { Amemset (NUM,0,sizeof(num)); thememset (TMP,0,sizeof(TMP)); + intTmax =0; - for(inti =0; I i) { $ intx; $scanf"%d", x); -++Tmp[x]; -Tmax =Max (Tmax, x); the } - for(inti =1; I

The main problem: solve the number of the factorial end of N 0.Analysis:There are 0 reasons for this: (1): 2 * 5 (2): 0 at the end of the multiplier, 10, 200However (2) can be attributed to (1), 10 = 2 * 5, 200 = 2 * 2 * 2 * 5 * 5It is easy to think of n! After decomposition into the product of prime number, only 2 * 5 This combination can produce the end of the 0, and 2 of the

Rightmost DigitProblem Descriptiongiven A positive integer N, you should output the most right digit of n^n.Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.Each test case is contains a single positive integer N (1Outputfor Each test case, you should output the rightmost digit of n^n. Sample Input234Sample OutputHintIn the first case, 3 * 3 * 3 = right