学步园

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;   
}