remainder theorem exercises

Theorem Description:Chinese remainder theorem: A method for solving congruence groups.For example, the following unary linear congruence equation Group:X≡A1 (mod M1)X≡A2 (mod m2)X≡A3 (mod m3). . . . . .X≡an (mod mn)Chinese remainder theorem: Assuming integers m1, M2, M3 ...,

Topic Link: http://acm.hdu.edu.cn/showproblem.php?pid=5768;Topic Analysis:Because it satisfies any group of pi and Ai, a "lucky number" can be "contaminated" and we could think of dealing with this problem through the repulsion. When we select a series of pi and AI, test instructions is converted to [x, Y] in order to be 7 divisible by 0, and by this series of Pi in addition to the number of AI, can be considered as a number of congruence equations of a single congruence equation set. Then we ca

Clear a misunderstandingAlthough the Chinese remainder theorem and the expansion of the Chinese remainder theorem is only two words, but they two of the solution difference of 108,000, so will not CRT does not matterUseLike $$\begin{cases}x \equiv b_{1}\pmod{a_{1}} \\x \equiv b_{2}\pmod{a_{2}} \\...\\x \equiv b_{n}\pmo

question is as follows: Any even number can be written as the sum of two prime numbers. Here, we will give you a number less than W. C will help you find the combination a + B = C, requiring a to be as small as possible, B is as big as possible. For million data, a prime number table is used directly. Poj 1006 portal poj 2891 Portal For more information about the Chinese Remainder Theorem, see my blog. HDU

The Chinese Remainder Theorem is used to solve x ≡ AI (mod mi). Among them, mi is mutually qualitative, and X has a unique solution. Let M be the product of MI, Wi = m/MI, WI about the module mi inverse is pi, that is, wi * PI + Mi * qi = 1. The above equations are equivalent to X W1 * P1 * A1 + W2 * P2 * A2 + ...... + wk * PK * ak (mod m ).............................. .................................. ①

Question address: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1573 Calculate the number of remaining Chinese understandings Note the Chinese Remainder theorem in the case of non-interconnectivity N = A1 (mod R1) N = a2 (mod R2) For example, n = A1 + R1 * x = a2 + R2 * Y.CubeCheng R1 * x-r2 * Y = a2-a1, Extended Euclidean. After solving X, we can see that n = A1 + R1 * X is a group of so

Combine two identical comodule equations at a time and then use the EXGCD solution.ll LCM (ll A,ll b) {returnA/__GCD (A, b) *b;}voidEXGCD (ll a,ll b,ll d,ll x,ll y) { if(b==0) {x=1; y=0;d =A; return; } exgcd (B,a%b,d,y,x); Y-=x* (A/b);} ll MLE (ll a,ll b,ll N) {ll x,y,d; EXGCD (A,n,d,x,y); if(b%d)return-1; X*= (b/d); return(x%n+n)%N;} ll Ex_china (LL*a,ll *m,intN) {ll x,y,d; ll M=1, a=0; REP (i,1, N) {ll k=mle (m[i],a-a[i],m); if(k==-1)return-1; A=k*m[i]+A[i]; M=LCM (M,m[i]); } re

1#include 2#include 3#include 4#include 5#include 6 #defineSC (x) scanf ("%d", x)7 #definePF (x) printf ("%d\n", X)8 #defineP printf ("\ n")9 #defineCL (x, y) memset (x.)Ten #definefor (I,X,V) for (int i=x; i One using namespacestd; A structnode - { - intnum; the stringstr; - }; - Const intMAX =5002; - intN, M, arr[Ten], Used[max], I, temp; +Queue Q; - voidBFS (); + intMain () A { at inti; - while(SC (N)! =EOF) - { - SC (M); -for (I,0, M-1) - SC (arr[i]); inSort (arr, arr+M)

HDU 1573 x problemTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 4857 Accepted Submission (s): 1611Problem 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 1 #defineM 152#include 3 using namespacestd;4#include 5#include 6 #definell Long Long7 voidEXGCD (ll a,ll b,ll x,ll y,ll gcd)8 {9 if(b==0)Ten { Onex=1; y=0;

BellTime limit:3000MS Memory Limit:32768KB 64bit IO Format:%i64d %i64u Submit Status Practice HDU 4767 Appoint Description:System Crawler (2015-03-15)Descriptionwhat? MMM is learning combinatorics!?Looks like she's playing with the bell sequence now:Bell[n] = number of ways to partition of the set {1, 2, 3, ..., n}e.g. N = 3:{1, 2, 3}{1} {2 3}{1, 2} {3}{1, 3} {2}{1} {2} {3}So, bell[3] = 5MMM wonders how to compute the bell sequence efficiently. Inputt--Number of casesFor each case:N (1 Bell[n]%

Topic PortalTest instructions: Very bare, just ask C (n, m)% (P1 * p2 * p3 * .... * PK)Analysis: First n,mHarvest: The wisdom of the ancestors of the crystallization must be learnedCode:/************************************************* author:running_time* Created time:2015/9/15 Tuesday 13:40:41* Fil E name:j.cpp ************************************************/#include 　　lucas+ China remainder theorem hdo

no solution then output-1.Sample Input2214 575 56519 54 40 24 8011 2 36 20 76Sample OutputCase 1:341case 2:5,996Authordigiter (Special Thanks Echo)Source2010 acm-icpc multi-university Training Contest (+)--host by bjtuRecommendzhouzeyong | We have carefully selected several similar problems for you:3573 3574 3575 3576 3577#include This problem trap: cannot output 0such as test data: for 23 40 0----->> should output Copyright NOTICE: This article for Bo Master original article, without Bo Master

China Remainder Theorem There is a question in Sun Tzu's Computing Theory: "Today there are things that don't know the number of things. There are two three, three, five, and three, and seven, and there are two. Are there things in ry ?" The current mathematical problem is X % 3 = 2, X % 5 = 3, X % 7 = 2, and returns the value of X; Many beginners of C language can't help but think of using brute force to c

.Sample Input28 711 9Sample Output31HintAll integers in the input and the output is non-negative and can is represented by 64-bit integral types.Test instructions: give you the K group number. X%m[i]=a[i]; thinking: China's residual theorem, expanding Euclid will not be able to refer to: http://blog.csdn.net/u010579068/article/details/45422941 reprint Please specify the Source: Find Star Child Topic Link: http://poj.org/problem?id=2891#include #defin

The ancients are very witty, but the Chinese remainder theorem for the divisor coprime is not able to solve the situation OH ~ ~ To use the previous method.Here is the detailed process of the Chinese remainder theorem:The last Step (3*5*7) represents the least common multiple of all the divisor. (because they coprime-.-)In my opinion ... The main idea of the Chin

One, test instructions: Chinese in upper right corner.Second, the idea:1, the equations are derived from test instructions2, using the Chinese remainder theorem to solve3, find the smallest positive integerThree, step:1, the equations are derived from test instructions(n+d)% (p);(n+d)% = e;(n+d)% = i;2, China remainder theore