linear congruence solver

that we get a first-to-last line. The starting point of setting Frog A is x, and Frog B's starting point coordinates are Y. Frog A can jump M m at a time, Frog B can jump n meters at a time, and two frogs will spend the same time jumping once. Latitude line total length l m. Now you have to find out how many times they have jumped before they meet.InputThe input includes only one line of 5 integer x,y,m,n,l, where X≠y OutputOutput the number of hops required to meet, and output a line of "impos

Topic Links:http://poj.org/problem?id=2115Main topic:For loop statements:for (int i = A; I! = B; i + = C)Statement 1;It is known that I, A, B, and C are all K-binary unsigned integer types, giving values of a, B, C, K, and calculating and outputting statements 1The number of executions, if infinite, then the direct output "FOREVER".Ideas:set the algorithm to perform X-steps, then the title becomes solution A + cx≡b (mod M) (m= 2^k). that is a + CX + my≡b. CX + MY ≡b-a (M = 2^k) is changed in or

Title: http://acm.hdu.edu.cn/showproblem.php?pid=3579Problem analysis: To find the smallest solution x of a linear congruence equation, it is important to note that if x equals 0, an integer interval LCM (a1,a2,a3,... An) is required to add a general solution to the equation group.There's nothing else to watch out for.#include #include#includestring.h>#include#includeusing namespacestd;__int64 a,b,c,d;__int

Topic Links:http://acm.hdu.edu.cn/showproblem.php?pid=3579Main topic:Kiki had x coins, she counted n times in different ways, each time she divided the coins into equal-sized groups and recorded each A group of coinsThe number of MI and the last remaining coins in the number of AI. So here's the question: how many coins are there in total?Ideas:The typical unary linear congruence equation Group x = Ai (mod

X problemTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 3850 Accepted Submission (s): 1228Problem 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

Title: http://poj.org/problem?id=2115The first two days were solved by a linear equation of two yuan, and the optimal solution was obtained by using the extended Euclidean original aim.Analysis:Number theory is not too much to understand, the solution of the time, met with no small trouble.Set the answer to X,n = (1IE (a+c*x) ≡b (mod n)//-----Results are obvious on both sides (A+CX)%n==bDegenerate C*x≡ (b-a) (mod n)//----with Comodule nature a-c==b-c

Defined: Pseudo-random number: A numerical sequence produced by a mathematical formula or algorithm. Although pseudo-random numbers are not random in mathematical sense, they can be used as true random numbers if they can be statistically verified. algorithm: The most basic idea of pseudo-random number generation is the uniform distribution (of course, this is not the only idea). Generally speaking, the random number function used in the mainstream programming language is basically using this

http://poj.org/problem?id=3087Set: S1={a1,a2,a3,... AC}S2={ac+1,ac+2,ac+3,.... A2C}TheTogether to becomeAc+1,a1,ac+2,a2 ... A2c,acAfter one conversion, it becomesS1={ac+1,a1,ac+2 ...}s2={... A2C,AC}Corresponds to the ordinal of each number that occurs before the change is+1,+2,+3....-c,-c+1,.....To think of the whole chain as a ring is also equivalent to:+1,+2,+3....+c,+c+1,.......A1, for example, must return to A1 after the a1->a2->a4->a7....c times.So the whole string passes a certain number o

http://acm.hdu.edu.cn/showproblem.php?pid=4686When you see n as a number less than 64-bit integer, you should have a feeling that this should be the fast power of the Tao Matrix in the ACM category.Ai,bi's recursive topic has been given,ai*bi=ax*bx* (ai-1*bi-1) +ax*by*ai-1+bx*ay*bi-1+ay*byAoD (n) =aod (n-1) +aibiConstruction Vector i{aod (i-1), ai*bi,ai,bi,1}Initial vector is i0={0,a0*b0,a0,b0,1}Structural Matrix a{1,0,0,0,0,1,ax*bx,0,0,0,0,ax*by,ax,0,0,0,bx*ay,0,bx,0,0,ay*by,ay,by,1,}Then i0* (

http://poj.org/problem?id=1426Tested, all values from 1-200 have a long long solution, so can be directly stored with a long longStarting from 1, each transfer to 10*s and 10*s+1, only the remainder can be stored,For the remainder I, only the first remainder of I is the most useful, only record this value can be#include 　　POJ 1426 Find The multiple idea, linear congruence, search difficulty: 2

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)Source acm-icpc multi-university Training Contest (--host by Bjtu)#include #include#includestring.h>#include#includeusing namespacestd;/*for X=r0 (mod M0) x=r1 (mod m1) ... x=rn (mod mn) input arrays M and array R, returns the 0,[m0,m1 of [mn]-1,..., x0] in the range that satisfies the above equation. All solutions for X are: X0+z*[m0,m1,... mn] (z is a

voidEXTENDGCD (Long LongALong LongBLong Longd,Long Longx,Long Longy) { if(b==0) {d=a;x=1; y=0;return;} EXTENDGCD (B,a%b,d,y,x); Y-= x* (A/b);}//solve A+c*x=b (mod D), return the smallest non-negative integer xLong LongModX (Long LongALong LongBLong LongCLong LongD) { if(a==B) {return 0; } if(c==0) { return-1; } Long Longx,y,tmpd; EXTENDGCD (C,d,tmpd,x,y); if((b-a)%tmpd! =0 ) { return-1; } Else{x*= (B-A)/tmpd; Long LongMoD = d/tmpd; X= (x%mod+mod)%MoD; return