中国剩余定理模板（2）出现不互质情况

来源：互联网发布：java程序员的晋升之路编辑：程序博客网时间：2024/05/18 01:11

蛋疼的说，如果出现不互质的情况，那就成解线性模余方程组了

神牛博客地址： http://yzmduncan.iteye.com/blog/1323599

/**  中国剩余定理（不互质）  */  #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;   }