Congruence equations
examples 1 : pku2891strange to Express integers
The number of the congruence equations for the remainder theorem in China is 22. However, the subject (the general situation, also includes 22 of the situation, so the Chinese remainder theorem becomes "The tears of The Times") MoD number may not be mutual, so we need to convert to ask.
P=B1 (mod a1); P/A1 ==?~~~~B1
P =b2 (mod a2);
P =b3 (mod A3);
......
P =bn (mod an);
The a1~an,b1~bn is given out.
Solution:
Article One: a1*x+b1= P
Article II: a2*y+b2= P
First strip minus second: A1*x-a2*y = b2-b1
Set A=A1,B=-A2,K=B2-B1, get X (actually call EXGCD when you ignore the minus sign in front of A2)
If k%d!=0, no solution
Otherwise, x=[(x* k/d)% (b/d) + (b/d)]% (b/d)
LCU says least common multiple
P= a1*x+b1+ several times LCU (A1,A2) (or y= (k-ax)/b, p=a2*y+b2+ several times LCU (A1,A2)
So the new b= a1*x+b1, the new a= LCU (A1,A2),
The new B as B1, the new A as A1, and then to A3 and B3 Union, until the end, the last new B is X
1#include <cstdio>2#include <cstring>3#include <cstring>4#include <iostream>5#include <algorithm>6#include <cmath>7 using namespacestd;8 #defineLL Long Long9 Ten LL a1,b1,a2,b2; One A LL Ax,ay; - ll EXGCD (ll A,ll b) - { the if(b==0) {ax=1, ay=0;returnA;} -LL G=EXGCD (b,a%b); -LL yy=ay; -ay=ax-a/b*ay;ax=yy; + returnG; - } + A intMain () at { -Freopen ("a.in","R", stdin); -Freopen ("a.out","W", stdout); - intN; - while(SCANF ("%d", &n)! =EOF) - { inscanf"%lld%lld",&a1,&B1); - BOOLok=1; to for(intI=2; i<=n;i++) + { -scanf"%lld%lld",&a2,&B2); the if(!ok)Continue; * LL a,b,c,g; $a=a1,b=a2,c=b2-B1;Panax Notoginsengg=EXGCD (A, b); - if(c%g!=0) {ok=0;Continue;} the if(b/g<0) b*=-1; +Ax= ((ax*c/g)% (b/g) + (b/g))% (b/g); Aa=b1+ax*A1; theg=a1*a2/EXGCD (A1,A2); +a1=g;b1=A; - } $ if(!ok) printf ("-1\n"); $ Elseprintf"%lld\n", B1); - } - return 0; the}
"poj2891"
2016-02-02 09:44:06
"poj2891" congruence equations