2016-05-23

Microcomputer principle: 8086 arithmetic instruction, jump instruction

Operating system: Process scheduling algorithm, banker algorithm

Python: Lists, dictionaries, classes and objects

Algorithm: Number theory (extended Euclidean, unary linear congruence equation set)

In the Hihocoder study these two, one dollar linear that question greatest common divisor variable uses to write disorderly, check code ability still need to strengthen ...

Number theory is very stressful.

Extended Euclid:

#include <stdio.h>#include <utility>using namespace std;typedef Long Long  ll;typedef pair<ll,ll>  xy;xy ans,cst (0,1 ); ll gcd (ll X,ll y) {if (x %y==0) return  y; else return gcd (y,x%  y);} XY EXTEND_GCD (ll a,ll b) {if (a%b==0) return  CST; XY TEMPXY=EXTEND_GCD (b,a%  b); xy x_y; x_y.first=  Tempxy. Second x_y.second=tempxy.first-a/b*  Tempxy.second; return  x_y;} int  Main () {ll s1,s2,v1,v2,m,a,b,c,judge; scanf ("%lld%lld%lld%lld%lld",&s1,&s2,&v1,&v2,&  m); a=v1-v2;b=m;c=s2- s1; if (a<0) a= (a+m)%  m; judge=  gcd (A, b); if (c%judge!=0) {printf (" -1\n"); return 0 ;} a/=  judge; b/=  judge; c/=  judge; ans=  EXTEND_GCD (A, b); ans.first= (ans.first*c)%  B; while ( ans.first<0) ans.first+=  b; printf ("%lld\n" , Ans.first); return 0 ;}       

One-element linear congruence Equation Group:

#include <stdio.h>using namespaceStd;typedef Long LongLl;structpair{ll x, y; Pair () {} pair (ll xx,ll yy) {x=xx;y=yy }};ll m[1010],r[1010];p Air ans,cst (0,1 ), LL gcd (ll X,ll y) {if (x%y==0) return  y; else return gcd (y,x%  y);} Pair EXTEND_GCD (ll A,ll b) {if (a%b==0) return  CST; pair TEMP=EXTEND_GCD (b,a%  b); pair xy; xy.x=  temp.y; x y.y=temp.x-a/b*  temp.y; return  XY;} int  Main () {ll n,i,r,m,a,b,c,judge; scanf ("%lld", &  n); for (i=1;i<=n;i++ ) scanf ("%lld%lld", &m[i],&  R[i]); R=R[1]; M=m[1 ]; pair ans; for (i=2;i<=n;i++ ) {a=m;b=m[i];c=r[i]- R; judge=  gcd (A, b); if (c%judge!=0) { printf (" -1\n"); return 0 ;} a/=judge;b/=judge;c/=  judge; ans=  EXTEND_GCD (A, b); ans.x= (c*ans.x)%  b; r=r+m*  ans.x; m=m/judge*  M[i]; r=r%  m;} while (r<0) r+=  M; printf ("%lld\n" , R); return 0 ;       

2016-05-23