remainder theorem exercises

by an integer value of zero. notice: For each integer in the input n (n Output Output n ^ n's digital root on a separate line of the output. Sample Input 2 4 0 Sample output 4 4 Author Eddy Give you a positive integer N and add all the numbers in N. If the result is less than 10, the result is the root number of n. If the result is greater than 10, then add the numbers in the above results. Now I will give you a number N, and calculate the number root of N ^ n. Idea: The data

HDU 6446 Unknown TreasureTest instructions: C (n, m)% (p[1] * p[2] P[k]) 0Idea: This problem is basically a template problem, Lucas theorem for C (n,m), and then use the Chinese remainder theorem combined with the model equation, because ll multiplies will cross the line, so use the bitwise multiplication.1#include 2#include 3#include 4#include 5#include 6#includ

1#include 2 using namespacestd;3typedefLong Longll;4 #defineLLD I64d5 #ifdef _WIN326 #defineLLD "%i64d"7 #else8 #defineLLD "%lld"9 #endifTen Const intN =10000+ -; One ll X,m,c,k; A intPos[n]; - /** - * Congruence theorem: the * (a+b) mod n = ((a mod n) + (b mod n)) mod n; - * AB mod n = (a mod n) * (b mod n) mod n; - * - *555 mod n = (550 + 5) mod n = 550 mod n + 5 mod n + * = ((mod n) * (ten mod n) + 5 mod n) mod n; - * g (i) mod n = (g (i-1) * + 5)

Title Link: Lucky7Test instructions: In the L and R range, satisfy the number of numbers that can be divisible by 7 and do not satisfy any group, x mod p[i] = a[i].Idea: tolerance theorem + Chinese remainder theorem + fast multiplication. (odd + even-)#include HDU 5768 China remainder

http://poj.org/problem?id=2891Test instructions: Solving a number x makes x%8 = 7,x%11 = 9; If x exists, the output is the smallest integer solution. otherwise output-1;Ps:Idea: This is not a simple question of China's residual theorem, since the input AI is not necessarily 22 coprime, while the condition of the Chinese remainder theorem is the divisor 22 coprime

NYOJ 420 p power summation (Fast Power + same remainder theorem) Description: A very simple question: 1 ^ p + 2 ^ p + 3 ^ p + ...... + The sum of n ^ p. Enter a separate number t in the first line to indicate the number of test data groups. Next, there will be t rows of numbers, each line includes two numbers n, p, Input 0 Output 1 ^ p + 2 ^ p + 3 ^ p + ...... + N ^ p returns the

; -m1=m2*m3; $M2=m1*m3; $M3=m1*m2; -y1=m1%M1; -y2=m2%m2; they3=m3%m3; -x= (a1*m1*y1+a2*m2*y2+a3*m3*y3)%M;Wuyi the returnx; - } Wu - About intMain () $ { - //x=2%3 x=a1%m1 - //x=3%5 x=a2%m2 - //x=2%7 x=a3%m3 A intm1=3, m2=5, m3=7; + inta1=2, a2=3, a3=2; theprintf"analytic method: \nx=%d \ n", Analytical (M1,M2,M3,A1,A2,A3)); -printf"Chinese remainder theorem: \nx=%d \ n", Chin

Title Address: http://acm.hdu.edu.cn/showproblem.php?pid=5446Idea: Lucas find out all A[i]=c (n,m)%m[i], the Chinese remainder theorem to find the final result X (Ll*ll will explode, handwritten multiplication).Chinese remainder theorem:Set m1,m2,....mn is a positive integer of 22 coprime, for any given integer A1 , a2,....an must have integers to satisfy x ≡a1 (

mod 5 = 4x MoD 7 = 6Han Xin is based on this equation set, the value of x is solved.Little hi: Well, that's it! We generalize this equation group to the general form: given n group Divisor m[i] and remainder r[i], the N group (M[i],r[i]) solves an X, making x mod m[i] = r[i]. Little ho: How do I feel that this equation set has a fixed solution?Small hi: This equation group is called the modular linear equations. It does have a fixed workaround. But b

3040 China remainder theorem 1time limit: 1 sspace limit: 32000 KBtitle Grade: Bronze BronzeTitle DescriptionDescriptionExcerpt from an Introduction to algorithms ...The number of the remainder is 2,3,2 when the first k is 3,5,7 removed;Enter a descriptionInput DescriptionA number k.Output descriptionOutput DescriptionFind the number of K qualifying.Sample inputS

24 8011 2 36 20 76Sample outputcase 1:341case 2:5,996Authordigiter (Special Thanks Echo)Source2010 acm-icpc multi-university Training Contest (+)--host by Bjtu idea, not coprime China remainder theorem, no solution output-1, has a solution to output a minimum positive integer (positive positive ...) Important things to say three times) that is the answer to the question of 0 when you need to output least c

Transmission DoorThinking of solving problemsExtension $CRT? $, that is, the Chinese remainder theorem in the case of the modulus is not coprime, first for the equation? $\begin{cases} x\equiv a_1\mod m_1\\x\equiv a_2\mod m_2\end{cases}$, it can be written as:$\begin{cases} x=k_1*m_1+a_1\\x=k_2*m_2+a_2\end{cases}$Then simultaneous equations:? $k _1*m_1+a_1=k_2*m_2+a_2$$\leftrightarrow-k_1*m_1+k_2*m_2=a_1-a_

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5768Give you n equations of congruence, and then give you l,r, and ask you how many numbers%7=0 and%ai! = Bi in L,r.The more obvious Chinese remainder theorem + tolerance, the time to add a (%7=0) this group.The middle will explode longlong, so add a fast multiplication (like a matrix fast power). Because the normal a*b is added directly to a B, it is li

Test instructions: number n, is (n+d)%23==p, (N+d)%28==e, (n+d)%33=i;Reprint please indicate source: http://www.cnblogs.com/dashuzhilin/;Idea: Chinese remainder theorem. Using the additive of congruence, the (n+d) is split into three number a,b,c,Make a%23==p,a%28==0,a%33==0;Make b%23==0,b%28==e,b%33==0;Make c%23==0,c%28==0,c%33==i;Then (n+d) = = (A+b+c) +LCM (23,28,33) *t;So, we can do the optimization, in

G- X problemTime limit:MS Memory Limit:32768KB 64bit IO Form At:%i64d %i64u SubmitStatusPracticeHDU 1573Appoint Description:System Crawler (2015-05-02)DescriptionHow many X are satisfied in a positive integer less than or equal to N: x mod a[0] = b[0], x mod a[1] = b[1], x mod a[2] = b[2], ..., x mod a[i] = b[i], ... (0 InputThe first behavior of the input data is a positive integer t, which indicates that there is a T group of test data. The first behavior of each set of test data is two

China Remainder Theorem To solve the following problems: For NMutual QualityReturns the smallest public solution x. x meets the requirements of X mod n1 = A1, X mod n2 = a2 ...... X mod nn = In our country, there is a mathematical ancient book "Sun Tzu's computing Sutra" with the following question: "Today there are things. I don't know the number. There are three, two, five, and three; seven to seven, two

Konjac Konjac does not have Chinese remainder theoremThere was a PE bug in the hand.Here is the AC code#include #includeusing namespacestd;intMain () {intp,e,i,d,t=1; CIN>>p>>e>>i>>D; Do{ intlcm=21252; intAns= (5544*p+14421*e+1288*I-D+LCM)%LCM; if(ans==0) ans=LCM; cout" Case"": The next triple peak occurs in""Days ."Endl; CIN>>p>>e>>i>>D; } while(p!=-1); return 0;}Ben Konjac Konjac don't know what this code meansBen Konjac Konjac is too weak.[

1 /*2 Congruence Equation Group:3 set positive integers m1.m2.mk 22, then equations4 X≡A1 (mod m1)5 x≡a2 (mod m2)6 x≡a3 (mod m3)7 .8 .9 X≡ak (mod mk)Ten There are integer solutions, One solution for X≡ (A1 * M1 * 1/m1 + A2 * M2 * 1/m2 + a3 * M3 * 1/m3 + ... +ak * Mk * 1/mk) mod M A WHERE M = M1 * M2 * M3 * ... * Mk, MI is M/mi, 1/mi is the inverse of Mi - */ - the voidEXGCD (intAintBintx,inty) - { - if(b = =0) { -x =1; +y =0; - returnA; + } AEXGCD (b, a%b, x, y); at intt =x;