number theory book

Link: HDU 5072 Coprime Given the number of N, ask how many 3 tuples can be selected, or [(a, B) = (B, c) = (a, c) = 1] or [(a, B) =1 and (a, c) =1 and (B, c) = 1]. Solution: You can change the angle of this question to think of three numbers as three sides of a triangle. The mutual quality means that the color of the side is 1. Otherwise, the value is 0. Number of triangles with the same color as the three

How many fibs? DescriptionRecall the definition of the Fibonacci numbers: f1 := 1 f2 := 2 fn := fN-1 + fN-2 (n>=3) Given two numbers a and B, calculate how many Fibonacci numbers are in the range [a, B]. InputThe input contains several test cases. each test case consists of two non-negative integer numbers A and B. input is terminated by a = B = 0. otherwise, a OutputFor each test case output on a single line the number of Fibonacci numbers fi

∵ Σ gcd (I, n) (1 = K1 * s (F1) + K2 * s (K2) +... + km * s (Km) {Ki is the approximate number of N, and S (Ki) is the condition that gcd (x, n) = KI (1 Export gcd (x, n) = KI (1 The number of X in gcd (x/ki, N/ki) = 1 (1 Radians = Σ PHI (N/ki) * ki ∴ O (SQRT (N) enumerative approx. 1 #include [Number Theory] [en

,n,d,x,y); - returnd==1? (x+n)%n:-1; - } - ll Log_mod (ll a,ll b,ll N) { -LL m,v,e=1, I; inM=SQRT (n+0.5); -v=Inv (POW (a,m,n), n); toMapMP; +mp[1]=0; - for(LL i=1; i) { theE= (e*a)%N; * if(!mp.count (e)) mp[e]=i; $ }Panax Notoginseng for(LL i=0; i) { - if(Mp.count (b))returni*m+Mp[b]; theb= (b*v)%N; + } A return-1; the } + - intMain () { $scanf"%lld%lld",t,k); $ while(t--) { -scanf"%lld%lld%lld",a,b,c); - if(k==1) { theprintf"%lld\n", pow

Title DescriptionDescriptionEnter two positive integer x0,y0 (2Condition: 1.p,q is a positive integer2. Require p,q to x0 for greatest common divisor, y0 as least common multiple.Trial: The number of all possible two positive integers that satisfy the condition.Enter a descriptionInput DescriptionTwo positive integers x0,y0Output descriptionOutput DescriptionThe number of all possible two positive integers

Two numbers if it's not coprime, then they must have a prime factor.Use VEC to deposit all prime-number factors from 2 to nWhether the number of the factor is read in by the VisOnly need to maintain the VIS array on the line when processing#include #include #include #include using namespace Std;const int MAXN = 100010;Vectorint ISP[MAXN];int VIS[MAXN];int NUM[MAXN];void Set (){memset (ISP, 0, sizeof (ISP));

DescriptionThere is a hill with n holes around. The holes is signed from 0 to N-1. A Rabbit must hide in one of the holes. A Wolf searches the rabbit in anticlockwise order. The first hole he get into was the one signed with 0. Then he'll get into the hole every m holes. For example, m=2 and n=6, the Wolf would get into the holes which is signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she'll survive. So we call these holes the safe holes.InputThe input starts with a

Test instructions: Asks if an integer is a prime number, if not, it outputs its smallest mass factor.Analysis:Determines whether a large integer is a prime number using the Miller_rabin algorithm, the POLLARD_RHO algorithm is used to find all factorization of a large integer. This problem is the direct set of templates.In addition the GCD and pow_mod here can not be used in the normal way, T. What I have co

Title Description DescriptionThe rational number in any [0,1] p/q (p, q are natural numbers) must be decomposed into 1/r1+1/r2+1/r3+...+1/rk, and R1The program requires finding the maximum decomposition of the p/q. input/output format input/output Input Format:Keyboard input P, q,1≤p≤q≤50output Format:From small to large in the output of each denominator in the decomposition, a line output a number input an

Codeforces Round #109 (Div. 1) B Number Theory// If two numbers are not mutually dependent, they must have a prime number factor.// Store all prime numbers of 2 to n using vec// Whether to read the number of factors stored in vis// You only need to maintain the vis array during processing.# Include # Include

HDU 4861 Couple doubi (number theory)HDU 4861 Couple doubi Question Link Given k and p, there are k balls. The value of each ball is 1 ^ I + 2 ^ I +... + (p-1) ^ I (mod p) (1 Idea: first-hand cannot lose, non-win is equal, so you only need to consider the value of each ball,By using the ferma's theorem or Euler's theorem, we can easily obtain that the cycle of this function is p-1, If I is a multiple of p-

Definition of prime number: Refers to the number of natural numbers greater than 1 that cannot be divisible by other natural numbers except 1 and the integer itself. 1 and 0 are not prime or composite. Prime numbers are two concepts that are relative to composite, and they form one of the most fundamental definitions of number

(Prime,0,sizeof(prime)); prime[1] = prime[0] =1; for(inti =2; I*i if(Prime[i])Continue; for(intj = i*i; J 1; }}intGCD (intAintb) {returnB?GCD (b,a%b): A;}intMain () {init (); while( ~scanf("%d%d", n, m)) {mp.clear (); for(inti =0; I scanf("%d", a[i]); for(inti =0; I scanf("%d", b[i]); g[0] = a[0]; for(inti =0; I true; for(inti = n1; I >=0; i--) {intx =0; for(intj =0; J intxx = x;intnum1,num2; NUM1 = Num2 =0; for(intj =2; J*j if(Prime[j])Continue;if(X%J)Continue;BOOLflag = Mp[j]; while(X%j = =0)

){ ++M[i][a]; N/=A; } ++A; } if(n>1) ++M[i][n]; } pri[0]=0; memset (Mark,0,sizeof(Mark)); for(intI=2; i -;++i) { if(!Mark[i]) pri[++pri[0]]=i; for(intj=1; j0]i*pri[j] -;++j) {Mark[i*pri[j]]=1; if(i%pri[j]==0) Break; } }}intFintNintk) {Mapint,int>MP; for(intI=2; ii) { intn=i; for(intj=1; j0]pri[j]j) { while(n%pri[j]==0){ ++Mp[pri[j]]; N/=Pri[j]; }}} Mapint,int>:: Iterator it; intans=1 -;

:#include #include#include#defineull unsigned long Longusing namespacestd;ull c[ the][ the];voidGet_c () {memset (C,0,sizeof(C)); c[1][0]=1; c[1][1]=1; for(intI=2; i A; i++) { for(intj=0; j) {C[i][j]=c[i-1][j]+c[i-1][j-1]; } }}intMain () {intN; intdata[ the]; intnum[Ten]; Get_c (); while(cin>>nN) {memset (num,0,sizeof(num)); for(intI=0; i) {cin>>Data[i]; Num[data[i]]++; } ull ans=0; for(intI=0; i9; i++) {ull k=0; if(Num[i]) {k=1; intm=n-1; Num[i]--; for(intj=0; j9; j + +)

of cats (the one in height) to clean the room, find out the number of cats that do not need to clean the room and the total height of all cats. Solution: Input H, W. Set K as the number of layers, Layer 1, Layer 2 ,.... Height: H/(n + 1) h/(n + 1) ^ 2 H/(n + 1) ^ 3 H/(n + 1) ^ 4 ...... H/(n + 1) ^ x = 1 Amount: 1 N ^ 2 N ^ 3 N ^ 4 ...... N ^ x = workers ① H * (1/(n + 1) ^ k = 1; → ③ H = (n + 1) ^ K; ② N ^

write a note to Dr Pincher detailing the proposed error checking mechanism noted. CRC Generation The message to was transmitted is viewed as a long positive binary number. The the message is treated as the most significant byte of the binary number. The second byte is the next most significant, etc. This binary number is called ' m ' (for message). Instead of

Test instructions: Given the initial value of the For loop, the end value and increment, and a modulo, the minimum number of cycles.Analysis:After reading the question should know is a concept of congruence, so it is to solve a one-time congruence equation, like the above problem with the expansion of Euclidean algorithm. The trick point of the problem is K Max is 32, then 2^32 out of int, to use a long long, so in 1Code:#include Copyright NOTICE: Th

Original link https://www.cnblogs.com/zhouzhendong/p/CF542D.htmlQuestion portal-cf542d question portal-51nod1477 Define the formula $ J (x) = \ sum _ {1 \ Leq k \ Leq X and k | X and \ gcd (K, X/K) = 1} k $. Given an integer $ A $, the number of positive integers $ x $ must be equal to $ J (x) = A $. $ X | N $ indicates that $ x $ is a factor of $ N $. $ \ Gcd (a, B) indicates the maximum common divisor of $ A $ and $ B $. $1 \ Leq A \ Leq 10 ^ {12} $

Order Status $ f (I, j) $ modulo $ d $ is $ I $, and is the minimum number for $ J $. You can use $ BFS $ to transfer But it's gone... Complexity $ O (10ds) $ #include Codeforces1070a find a number Graph Theory