number theory book

HDU 1212 Topic Link Click to open linkDescription: Given a large number a, the result of modulo B is obtained.Problem analysis: Because A is large, you need to introduce a string for processing!Algorithm Analysis: Congruence theorem1. (m + N)% c = (m% c + n% c)% c2. (m* N)% c = ((m% c) * (n% c))% c3. (m ^ n)% c = ((m% c) ^ n)% c (this theorem can be used for fast power calculation for further discussion)This problem requires the use of Theorem 1. For

inferred prime is not an approximate number of n---in fact, with the help of a pair of approximate, small always complexity of time. )2, suppose at first glance did not find the law or no train of thought. Simulate several numbers yourself. A continuous simulation. not limited to sample input. Find the rules for yourself .int Prmcnt;bool is[n];int prm[m];int getprm (int N) { int i,j,k=0; int s,e= (int) (sqrt (0.0+n) +1); CL (is,1); prm[k

]) $ // {Panax Notoginseng //cout - intsum =0;BOOLOK =true; the for(intj =1; J ) + { A intTMP = i/J; the //cout + if(Sum > S) {OK =false; Break;} - if(Tmp*j! = i)Continue; $ if(tmp! = j) Sum + =J; $Sum + =tmp; - } - if(sum = = S) {exist =true; Break;} the // } - }Wuyi

159 #defineS 500010//pay attention to the problem when 5*1e5Ten#include One using namespacestd; A#include -#include -typedefLong Longll; the ll L,r; - BOOLvis[s]={0}; - intN,mn= (1 to)-1, a[n]; - ll Dis[s]; + voidinput () - { +Cin>>n>>l>>R; A for(intI=1; ii) at { -scanf"%d",a[i]); - if(a[i]==0) - { -i--;n--; - Continue; in } -mn=min (mn,a[i]); to } + } - voidSPFA () the { *queueint>Q; $Q.push (0);Panax Notoginsengvis[0]=true; -dis[0]=0;/*Not

the X1+X2+X3+...XN = x solution of the group number, x∈[0,m].Analysis: First we face the absence of empty trees, using the basic partition principle (), easy to get C (m-1,n) group, which is obvious. However, the crux of the problem is to allow the existence of empty trees, so we need to select the number of empty trees, that is, in the selected m-1 elements and add n trees, in which the selection of n-1 e

FibonacciTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 3654 Accepted Submission (s): 1671Problem Description arrived in 2007. After 2006 years of cultivation, mathematical prodigy Zouyu finally put 0 to 100000000 of Fibonacci seriesThe values (f[0]=0,f[1]=1;f[i] = f[i-1]+f[i-2] (i>=2) are all backed down.Next, Codestar decided to test him, so each asked him a number, he will say the answer, but som

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)