Han Xin and His Troops

Han Xin and His Troops

中国剩余定理

JAVA板子

/*中国剩余定理，根据公式需要求取大数的逆元*/
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {

        static BigInteger m[]=new BigInteger[105];
        static BigInteger c[]=new BigInteger[105];
        static int  n;
        static BigInteger _m;
        static BigInteger x,y;
        public  static BigInteger[] ex_gcd(BigInteger a,BigInteger b){
            BigInteger ans;
            BigInteger[] result=new BigInteger[3];
            if(b.equals(BigInteger.ZERO))
            {
                result[0]=a;
                result[1]=BigInteger.ONE;
                result[2]=BigInteger.ZERO;
                return result;
            }
            BigInteger [] temp=ex_gcd(b,a.mod(b));
            ans = temp[0];
            result[0]=ans;
            result[1]=temp[2];
            result[2]=temp[1].subtract((a.divide(b)).multiply(temp[2]));
            //System.out.println(result[0]+" "+result[1]+" "+result[2]);
            return result;
        }
        static BigInteger getInv(BigInteger a,BigInteger md)
        {
            //x=BigInteger.ZERO;y=BigInteger.ZERO;
            BigInteger d[]=ex_gcd(a,md);
            //System.out.println(a+" "+md);
            //System.out.println(x+" "+y);
            //System.out.println(d[1]);
            return (d[0].equals(BigInteger.ONE))?(d[1].add(md)).mod(md):BigInteger.valueOf(-1);
        }
        static BigInteger exCRT(int n)
        {
            BigInteger m1,m2,c1,c2,d;
            for(int i=2;i<=n;i++)
            {
                m1=m[i-1];m2=m[i];c1=c[i-1];c2=c[i];
                d=m1.gcd(m2);
                if(!(c2.subtract(c1)).mod(d).equals(BigInteger.ZERO))return BigInteger.valueOf(-1);//此时无法合并
                m[i] = m[i-1] .multiply(m[i]).divide(d) ;
                BigInteger t=(c2.subtract(c1)).divide(d);
                //System.out.println(t);
                c[i] = t.multiply( getInv(m1.divide(d),m2.divide(d))) .mod ( m2.divide(d) ).multiply(m1) .add(c1) ;
                //System.out.println(c[i]);
               // System.out.println(getInv(m1.divide(d),m2.divide(d)));
                c[i] = ( (c[i] .mod(m[i])) .add(m[i]) ) .mod( m[i]);
               // System.out.println(m[i]);
            }
            return c[n];
        }
     public static void main(String[] args) {

            Scanner sc=new Scanner(System.in);
            n=sc.nextInt();
            _m=sc.nextBigInteger();
           // int k=0;
            for(int i=1;i<=n;i++){
                m[i]=sc.nextBigInteger();
                c[i]=sc.nextBigInteger();

            }

            BigInteger ans=exCRT(n);
            //System.out.println(ans);
            if(ans.compareTo(_m)>0){
                System.out.println("he was probably lying");
            }else if(ans.equals(BigInteger.valueOf(-1))){
                System.out.println("he was definitely lying");
            }else System.out.println(ans);

        }

}

原文地址：https://www.cnblogs.com/liulex/p/11370275.html