best number theory books

that size. Sample Input 4245231 Sample output 1 2 52 4 133 5 214 231 32549 SourceGreater New York 2006 Mean: In the first quadrant, enter N, and then count the number of viewpoints in the range of (0 The so-called viewpoint can be reached (x1, Y1) starting from (0, 0) without any point of intersection in the middle. Analyze: Through analysis, we will find that as long as X and Y are mutually qualitative, (x, y) is the viewpoint. We only need to o

minimum time (in seconds).If Dr Hanks chooses which cell does not satisfy the requirement, the output integer-1.1#include 2#include 3 structPRM4 {5 intx,t;6}m[10010];7 inta[10010];8 intMain ()9 {Ten inti,j,k,n,p,q,m1,m2,x,y,z,cntm,ans,t; One BOOLOK; Ascanf"%d",n); -scanf"%d%d",m1,m2); -cntm=0; the for(i=2; m1!=1; i++) - if(m1%i==0) - { -cntm++; +m[cntm].x=i; - while(m1%i==0) + { Am[cntm].t++; atM1/=i; - } -m[cntm].t*=m2; - } -ans=-

;Ideas:Note that the number is within 1000, so the combination of all to find out, but the number is huge so the same to be compressed;AC Code:#include #include#include#includeusing namespaceStd;typedefLong Longll;ll dp[1003],num[1003],p[1003][1003];intn,a[1003];Constll mod=1e8+7;//Note is 8, the time of the game this place direct wa to cryintgcdintXinty) { if(y==0)returnx; returnGCD (y,x%y);}intMain ()

Question: There are N nodes in the circle, and a node in the center is connected to N nodes. There are 2 * N edges in total and N edges are deleted, to connect N + 1 points, the same rotation is considered equivalent. How many situations are there.Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2481It is said that there was only one Tsinghua team in the competition. Very comprehensive, mainly because the recursive part is hard to think about.Good questions, difficult !!!!!!!!!!!Practice from AC

Topic Links:option=com_onlinejudgeitemid=8page=show_problemproblem=1253 ">10312-expression Bracketing Test instructions: There are n x, which requires parentheses to infer the number of non-binary expressions. train of thought: Two the expression of the fork is equal to the number of Catalan, then only the total number is calculated, minus the

large binary numbers exist in the array, such as the conversion into a binary system, first preserved, and then multiplied by 1000 or 10,000, the top of the for is relative to the decimal, the now*10 to the corresponding binary can be changed, The remainder of the enumerator only need to print the prime number of the table, a start afraid K too big can not go down and not to be directly converted into million, the result has been WA, and later conver

Codeforces Round #257 (Div. 2) E (number theory + structure)E. Jzzhu and Applestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output Jzzhu has pickedNApples from his big apple tree. All the apples are numbered from 1N. Now he wants to have them to an apple store. Jzzhu will pack his apples into groups and then them. each group must contain two apples, and the

Mean: Description: An array containing non-negative integers (N in length) is used to find the maximum multiple of 3 consisting of these numbers. If not, impossible is output. Analyze: The first thing that comes to mind is direct violence. This is the dumbest method. If there is a large amount of data, it must be TLE. Directly use brute force to generate all the combinations, which are 2 ^ N. Compare and judge each number. It takes O (n) time, because

,Y0) General Solution: x=x0+ (b/d) K, y=y0-(A/DK) is to try to get the positive and negative mutual cancellation, ( true • Primary School Olympiad content)Doubts: Most beginners may have such a question (anyway I have just learned), to solve the equation is ax+by=m, and the above algorithm is the solution of AX+BY=GCD (A, B)In fact, AX+BY=GCD (A, b) can become AX*K+BY*K=GCD (A, b) *k, so that gcd (a, b) *k=m, find K, and then X*k is the solution of ax+by=m. 5 , multiplication inverse elementLe

) { if(!Prime[i]) {PO[I].A[PO[I].M++]=i; for(intj=2*i;ji) {po[j].a[po[j].m++]=i; PRIME[J]=1; }}}}ll Check (intx) {LL s=1; for(intI=1; i) {s*=2; } returns;}intGetans (intXinty) { intL=0, R= (int) (Log (y/x+1)/log (2.0))+1; while(lr) {intMid= (l+r) >>1; if(Check (mid) *x>=y) r=mid-1; ElseL=mid+1; } returnr+1;}voidSolveintn,ll m) { if(n==1) { if(m==1) printf ("0\n"); Elseprintf"-1\n"); } Else { intans=0;

to) a hole to catch the rabbit, ask the rabbit finally can escape a robbery. For example, there are n=6 a hole, the number is 0, 1, 2, 3, 4, 5, the wolf every m-1=1 a hole to catch the rabbit, then the wolf into the cave is 0, 2, 4, 0, 2, 4 ... (infinite loop). If rabbits hide in holes 1, 3 and 5th, they are safe.Sample Input2//Number of test cases1 2//m and N2 2Sample OutputNOYES#include using namespacest

; I i) {if(!bz[i]) pri[++cnt] = i, phi[i] = i-1; for(intj =1; J j) {if(i * pri[j] > N) Break; Bz[i* Pri[j]] =1; if(i% pri[j]! =0) Phi[i * Pri[j]] = phi[i] * (Pri[j]-1); if(i% pri[j] = =0) {Phi[i* Pri[j]] = phi[i] *Pri[j]; Break; } }}11. Number of combinations(1) C (m, n) = m! /n! (M-N)!Example: NOIP2016 d2t1 Combinatorial number problemSolution: Yang Hui triangle pretreatment, the two-dimensional prefi

Topic Links:Codeforces 201AMain topic:Give an X, to find a square matrix of the smallest side length, the matrix is 01 matrix, and meet the elements up and down symmetry, left and right symmetry, asked to construct the number of 1 is the smallest matrix of x side length is how much.Topic Analysis: First we can see that if the maximum number of n-1 constructs is greater than the maximum

UV-575-Skew Binary (simple number theory !) Uvs-575 Skew Binary Time Limit:3000 MS Memory Limit:Unknown 64bit IO Format:% Lld % llu Submit StatusDescription Skew Binary When a number is expressed in decimal,K-Th digit repre

can divide n+1 maximum number of times).That is, the product of all P^maxn divided by the product of p^K, the numerator equals [1, 2, 3,.... n+1], and the denominator equals n+1. The result is consistent with the conclusion of the puzzle.Prove the process a little rubbing ....If K and maxn are found in order, the complexity is O (Num*log (p)), where num is the number of primes ,P is the prime

-];intnum[107][1 -];intprime[]={2,3,5,7, One, -, -, +, at, in, to,Panax Notoginseng, the, A, +, -, -};intstate[ -];voidPrint (intIintu) {if(i = =1) {printf("%d", Num[i][u]);return; } Print (I-1, Pre[i][u]);printf("%d", Num[i][u]);}intMain () { while( ~scanf("%d", n)) { for(inti =1; I scanf("%d", a[i]);memset(DP,0x3f,sizeof(DP));intINF = dp[0][0]; dp[0][0] =0;memset(State,0,sizeof(state)); for(inti =2; I -; i + +) for(intj =0; J -; J + +)if(I%prime[j] = =0) State[i] |= (1intTotal =1 -; for

issues related to the sum, he counted the number of approximate numbers of each positive n and expressed it in F (n). Now Xiao Lian hopes to use "Samuel2" to count the sum of f (1) to F (N) and M.F (n) represents an approximate number of N, which is now given N, which requires the sum of f (1) to F (n) to be calculated.input/output formatInput format:Enter a line, an integer nOutput format:Outputs an integ

Divisible:　　introduced the \ Representative division of. M\n indicates that M is divisible by N. Pay attention to the divide here. Represents n = km (k is an integer).In the divisible sex here. M must be a positive number. Perhaps you can describe N as the K-fold of M. In this description, M can exactly be any number. And in the division of the expression m divisible by N, which specifies that M must be a p

: gives A,b,c,d,k, which makes a1 Set F (k) for gcd (x, y) =k number pairs (x, y) logarithm, we require f (1)2 set F (k) is gcd (x, y) is a multiple of the number of K (x, y) logarithm, can think of f (k) = Floor (b/k) *floor (d/k), F (k) =e[b/k]*[d/k]*mu[k];3 the Möbius was reversed by: 4 Order lim=min (b/k,d/k)5 F (1) =mu[1]*f (1) + mu[2 ]*f[2] + ... + mu[lim]*F (Lim)6 because (N1,N2) and (N2,N1) are

1435. Mutual Quality Time Limit: 1000 MSMemorylimit: 65536 KTotal submissions: 126 (48 Users)Accepted: 30 (26 Users)[Mysolution] Description Euler's function plays an important role in number theory. The value of an Euler's function of a number represents the number of positive integers that are mutually compatible wit