remainder theorem exercises

P1164Cao Chong Pig accepted label: Three Kingdoms hegemony [show label]DescribeSince Cao Chong finished the elephant, Caocao began to fathom let son do some career, so sent him to Zhongyuan pig farm pig, but Cao Chong full not happy, so in the work so-so, there was a time when Caocao want to know the number of sows, so Cao Chong want to play Caocao a. For example, if there are 16 sows, if 3 pigsty are built, there is no place for the 1 pigs left. If you build 5 pigsty, but there are still 1 pigs

Title 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 inputSample Input1Sample outputSample Output23Data range and TipsData Size Hintk>=1;The answer is no more than the range that long long can store.A typical number theory problem.// Chines

> Analysis>> Three Cycles is three Inma, can be very simple to use the Chinese remainder theorem> attached Code1 /* -------------------------2 * Chinese remainder theorem3 * -------------------------*/4#include"stdio.h"5 6 intMainvoid)7 {8 intp =0, E =0, i =0, d =0 ;9 intDays =0 ;Ten intCount =0 ; One A while(1) - { -scanf"%d %d%d%d", p, e,

Euler's Function Chinese Remainder Theorem in algorithm Summary 1. Euler's Function 　　Concept:In number theory, for a positive integer N, the Euler's function is the number of numbers that are less than or equal to the number of numbers that intersect with N. 　　General Formula: Phi (x) = x (1-1/P1) (1-1/P2) (1-1/P3) (1-1/P4 )..... (1-1/PN) 　　　　P1, P2 ...... All prime factors whose PN is x. x is an integer

HDU 1573 X (China Remainder Theorem) [Question]: calculate the number of X values 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 [Idea] The application of China's residual theorem is to find the number that meets certain conditions. Code: /** Problem: HDU No.1573 * Runnin

Exgcd Extended Euclidean void exgcd(int a,int b,int x,int y){ if(!b){x=1,y=0;return;} exgcd(b,a%b,x,y);b-=y*(a/b);} For \ (gcd (a, B) = g \), \ (A \ times K_1 + B \ times K_2 = g \) Run \ (exgcd (a, B, x, y) \ (k_1 = x + k \ times B \) For \ (gcd (a, B) = g \), \ (A \ times K_1 + B \ times K_2 = C \ times g \) Run \ (exgcd (a, B, x, y) \ (k_1 = x \ times C + k \ times B \)China Remainder Theorem Here

Topic linksThe value of C (n, m)%p, N, mFirst, the value of the C (n, m)%PI is calculated, and then this is an equation of congruence. Solved by the Chinese remainder theorem.#include #include#include#include#includeusing namespacestd;#definell Long Longll a[ -], b[ -];voidExtend_euclid (ll A, ll B, ll x, LL y) { if(b = =0) {x=1; Y=0; return; } extend_euclid (b, a%b, x, y); LL TMP=x; X=y; Y= tmp-(A/b) *y

Information reference: https://zh.wikipedia.org/wiki/%E4%B8%AD%E5%9B%BD%E5%89%A9%E4%BD%99%E5%AE%9A%E7%90%86"There is no known number, 33 of the remaining two, 55 of the remaining three, 77 of the remaining two." Asking for geometry? "This is the source of the Chinese remainder theorem, what does it mean?" That is, if the number of items, if three three number, there will be two, if five five numbers, there

http://acm.hdu.edu.cn/showproblem.php?pid=1573X problemTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 4439 Accepted Submission (s): 1435Problem description The number of x in a positive integer less than or equal to n is satisfied: 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 The first behavior of input data is a positive integer t, which indicates that there is a T group of test data. The first behav

Topic Link: Click to open the linkThe main topic: there is a number X,x%ai = ri, give n to Ai and ri, ask what is the smallest nonnegative integer of x, if there is no output-1The Chinese remainder theorem of non-reciprocity:x%a1= R1; X%A2 = R2; Set K1,K2 to get x = A1*k1 + r1, x = a2*k2+r2So a1*k1+r1 = a2*k2+r2--and A1*K1 = (R2-R1) + A2*K2----> a2 the entire equation, get (a1*k1)%a2 = (R2-R1)%a2, there is

x ProblemTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 3921 Accepted Submission (s): 1253Problem description The number of x in a positive integer less than or equal to n is satisfied: 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 The first behavior of 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 positive i

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

China remainder theorem coprime editionThe M1,M2,M3,...,MK is a positive integer of 22 biotin, i.e. gcd (MI,MJ) =1,i!=j,i,j=1,2,3,..., K.Then the same residual equation group:x = A1 (mod n1)x = A2 (mod n2)...x = AK (mod nk)modulo [N1,n2,... NK] has a unique solution, that is, in the sense of [N1,n2,..., NK], there is a unique x that satisfies:x = AI mod [n1,n2,..., nk], i=1,2,3,..., K.The solution can be wr

unluckily they haven ' t found the height of T He bamboo. To is more exact, they knowN, all pi and corresponding ri, but not L. So, you came forward and found the task challenging and so, you want to find L, from the given information.InputInput starts with an integer T (≤10000), denoting the number of test cases.Each case is starts with a line containing an integer n (1≤n≤12). Each of the next n lines contains integers pi (1 ri (0 . All pi would be distinct.OutputFor each case, print the case

1 Chinese remainder theorem:Set positive integer m1, M2, M3. mk 22, then a congruence equation Group X≡ai (mod mi) i = 1, 2, 3, ..., K has an integer solution, and in mod m = m1* m2 * M3 ... mk solution is unique, that is, any two solution is mod m with the remainder OfSet mi = m/mi; SoBecause mi (i = 1, 2, 3, ...), I! = k) is a multiple of MK, which can go around,While theWhich is the inverse, so ≡1 (mod m

Strange Way to Express Integers (Chinese Remainder Theorem + non-interconnectivity), strangeintegersStrange Way to Express IntegersTime Limit:1000 MSMemory Limit:131072KB64bit IO Format:% I64d % I64uSubmit Status Practice POJ 2891 Appoint description:System Crawler) Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is describe