EXTEND_GCD solving indefinite equation/membrane linear equation/multiplication Inverse element

Shaped like A*x+b*y=c

For indefinite equations, a,b>0 actually doesn't matter, because GCD (A, B) =gcd (|a|,|b|) GCD to greatest common divisor

Known by the theorem of number theory, when c%gcd==0, indefinite equations have solutions, now we seek this solution.

GCD=GCD (A, b); A*B=GCD*LCM; LCM for Least common multiple

A ' =a/gcd;b ' =b/gcd;c ' =c/gcd; A ' currently with B ' coprime

Make C ' =1=a ' *x+b ' *y;

A set of special solutions is obtained by using EXTEND_GCD. (X0,Y0)

(X0*c '. Y0*c ') a set of special solutions for a ' *x+b ' *y=c '

The solution of the equation is to find out its general solution; X=x0*c ' +b ' *t; Y=y0*c '-a ' *t; solve the problem with the general solution

It took a whole night to tangle up an obvious question today. Sure enough, lookers.

I want to laugh now. haha haha haha

uva,10413

The title means: T a case,n Savage, c,p,l respectively represents the initial cave, the next day to move the number of caves, survival days.

Q: The smallest number of caves, which makes the N savages not met in their respective lifetimes.

Data: 1*10^6 can be accepted, so this problem with the initial value of the savage in the original maximum number of caves, enumeration can be over.

Source:

1#include <iostream>2#include <algorithm>3#include <cstdio>4 5 using namespacestd;6 7 intX,y,gcd,n;8 intc[ -],p[ -],l[ -];9 Ten voidEXTEND_GCD (intAintb) One { A     if(b==0) -     { -x=1; they=0; -Gcd=A; -         return; -     } +EXTEND_GCD (b,a%b); -     inttmp=x; +x=y; Ay=tmp-a/b*y; at     return; - } -  - BOOLSolveintIintJintm) - { -     intc=c[i]-C[j]; in     intb=m; -     inta=p[j]-P[i]; to EXTEND_GCD (A, b); +     if(C%GCD)return false; -x*=c/gcd; the     if(gcd<0) gcd=-gcd; * //because of the X=x+b/gcd*k, X contains a number of algebraic and differential b/gcd $ //so take the X to the B/GCD and pay attention to the minimum positive value of the negative number x.Panax Notoginsengx= (x (B/GCD) +b/gcd)% (b/gcd); -     if(X<=l[i] && x<=L[j]) the         return true; +     return false; A } the BOOLTestintm) + { -      for(intI=0; i<n-1; i++) $          for(intj=i+1; j<n;j++) $         { -             if(Solve (I,J,M))return false; -         } the     return true; - }Wuyi  the  - intMain () Wu { -     intT; Aboutscanf"%d",&t); $      while(t--) -     { -scanf"%d",&n); -         intlb=0; A          for(intI=0; i<n;i++) +         { thescanf" %d%d%d", c+i,p+i,l+i); -lb=Max (lb,c[i]); $c[i]--; the         } the          while(lb<=1000000) the         { the             if(Test (LB)) Break; -lb++; in         } theprintf"%d\n", LB); the     } About     return 0; the } the //provide test data the Sample Input + 2 - 3 the 1 3 4Bayi 2 7 3 the 3 2 1 the 5 - 1 2  - - 4 4 6 the 8 5 9 the  One 8  - the  - 9 Ten the Sample Output - 6 the  -

EXTEND_GCD solving indefinite equation/membrane linear equation/multiplication Inverse element