欧几里得-GCD及扩展

GCD最大公约数

int gcd(int x,int y)
{
    if(!x || !y) return x>y?x:y;
    for(int t;t=x%y;x=y,y=t);
    return y;
}

快速GCD

int kgcd(int a,int b)
{
    if(a==0) return b;
    if(b==0) return a;
    if(!(a & 1) && !(b & 1)) return kgcd(a>>1,b>>1)<<1;
    else if(!(b & 1)) return kgcd(a,b>>1);
    else if(!(a & 1)) return kgcd(a>>1,b);
    else return kgcd(abs(a-b),min(a,b));
}

扩展GCD

求x,y使得gcd(a,b)=a*x+b*y;

int extgcd(int a,int b,int &x,int &y)
{
    if(b==0) {x=1;y=0;return a;}
    int d=extgcd(b,a%b,x,y);
    int t=x;
    x=y;
    y=t-a/b*y;
    return d;
}

模线性方程 a*x=b(%n)

void modeq(int a,int b,int n) //! n>0
{
    int e,i,d,x,y;
    d=extgcd(a,n,x,y);
    if(b%d>0) printf("No answer!\n");
    else 
    {
        e=(x*(b/d))%n;
        for(i=0;i<d;i++) //!!! here x maybe<0
        {
            printf("%d-th ans:%d\n",i+1,(e+i*(n/d))%n);
        }
    }
}

模线性方程组

a=b[1](%w[1]);a=b[2](%w[2]);...a=b[k](%w[k]);

其中w,b已知,w[i]>0且w[i]与w[j]互质，求a;(中国剩余定理)

int china(int b[],int w[],int k)
{
    int i,d,x,y,m,a=0,n=1;
    for(i=0;i<k;i++) n*=w[i];
    for(i=0;i<k;i++)
    {
        m=n/w[i];
        d=extgcd(w[i],m,x,y);
        a=(a+y*m*b[i])%n;
    }
    if(a>0) return a;
    else return (a+n);
}

不要求w[i]与w[j]互质;

#include <iostream>  
#include <cstdio>  
#include <cstring>  
using namespace std;  
typedef __int64 int64;  
int64 Mod;  
int64 gcd(int64 a, int64 b)  
{  
    if(b==0)  
            return a;  
        return gcd(b,a%b);  
    }  
      
    int64 Extend_Euclid(int64 a, int64 b, int64&x, int64& y)  
    {  
        if(b==0)  
        {  
            x=1,y=0;  
            return a;  
        }  
        int64 d = Extend_Euclid(b,a%b,x,y);  
        int64 t = x;  
        x = y;  
        y = t - a/b*y;  
        return d;  
    }  
      
    //a在模n乘法下的逆元，没有则返回-1  
    int64 inv(int64 a, int64 n)  
    {  
        int64 x,y;  
        int64 t = Extend_Euclid(a,n,x,y);  
        if(t != 1)  
            return -1;  
        return (x%n+n)%n;  
    }  
      
    //将两个方程合并为一个  
    bool merge(int64 a1, int64 n1, int64 a2, int64 n2, int64& a3, int64& n3)  
    {  
        int64 d = gcd(n1,n2);  
        int64 c = a2-a1;  
        if(c%d)  
            return false;  
        c = (c%n2+n2)%n2;  
        c /= d; 
        n1 /= d;  
        n2 /= d;  
        c *= inv(n1,n2);  
        c %= n2;  
        c *= n1*d;  
        c += a1;  
        n3 = n1*n2*d;  
        a3 = (c%n3+n3)%n3;  
        return true;  
    }  //求模线性方程组x=ai(mod ni),ni可以不互质  
    int64 China_Reminder2(int len, int64* a, int64* n)  
    {  
        int64 a1=a[0],n1=n[0];  
        int64 a2,n2;  
        for(int i = 1; i < len; i++)  
        {  
            int64 aa,nn;  
            a2 = a[i],n2=n[i];  
            if(!merge(a1,n1,a2,n2,aa,nn))  
                return -1;  
            a1 = aa;  
            n1 = nn;  
        }  
        Mod = n1;  
        return (a1%n1+n1)%n1;  
    }  
    int64 a[1000],b[1000];  
    int main()  
    {  
        int i;  
        int k;  
        while(scanf("%d",&k)!=EOF)  
        {  
            for(i = 0; i < k; i++)  
                scanf("%I64d %I64d",&a[i],&b[i]);  
            printf("%I64d\n",China_Reminder2(k,b,a));  
        }  
        return 0;  
    }